diff --git "a/all/mathqa_probability.json" "b/all/mathqa_probability.json" new file mode 100644--- /dev/null +++ "b/all/mathqa_probability.json" @@ -0,0 +1,4896 @@ +{ + "Source": "https://arxiv.org/pdf/2108.07732.pdf", + "Categories": [ + { + "Math complexity": 6, + "Language complexity": 7, + "Domain knowledge complexity": 5 + } + ], + "Instances": [ + { + "Input": "a gardener is going to plant 2 red rosebushes and 2 white rosebushes . if the gardener is to select each of the bushes at random , one at a time , and plant them in a row , what is the probability that the 2 rosebushes in the middle of the row will be the red rosebushes ?", + "Output Program": [ + "import scipy\nn0 = 2.0\nn1 = 2.0\nn2 = 2.0\nt0 = n0 + n0\nt1 = scipy.special.comb(t0, n0)\nanswer = 1.0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.16666666666666666" + ], + "split": "train" + }, + { + "Input": "in a throw of dice what is the probability of ge \u00e6 \u00ab ng number greater than 3", + "Output Program": [ + "n0 = 3.0\nt0 = 6.0 - n0\nanswer = t0 / 6.0\nprint(answer)" + ], + "Output Answer": [ + "0.5" + ], + "split": "train" + }, + { + "Input": "given a bag with 2 red balls and 3 yellow balls , you randomly select one of them , in order . if the second ball you take is yellow , what is the probability that the first one was also yellow ?", + "Output Program": [ + "n0 = 2.0\nn1 = 3.0\nt0 = n0 + n1\nt1 = t0 - 1.0\nanswer = n0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.5" + ], + "split": "train" + }, + { + "Input": "a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability that out of the two , one is white and one is blue ?", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n0 + n1\nt1 = n1 + 2.0\nt2 = n0 * n1\nt3 = t0 * t1\nt4 = t3 / 2.0\nanswer = t2 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.5357142857142857" + ], + "split": "train" + }, + { + "Input": "a coin is tossed 4 times . what is the probability of getting exactly 3 heads ?", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 3.0\nt0 = scipy.special.comb(4.0, 3.0)\nt1 = 1.0 / 2.0\nt2 = t0 * t1\nt3 = t1**min(n1, 5)\nanswer = t2 * t3\nprint(answer)" + ], + "Output Answer": [ + "0.25" + ], + "split": "train" + }, + { + "Input": "if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is -", + "Output Program": [ + "t0 = 3.0 / 52.0\nt1 = 52.0 / 4.0\nt2 = t1 / 52.0\nanswer = t0 + t2\nprint(answer)" + ], + "Output Answer": [ + "0.3076923076923077" + ], + "split": "train" + }, + { + "Input": "what is the probability that the sum of two dice will yield a 9 , and then when both are thrown again , their sum will again yield a 9 ? assume that each die has 8 sides with faces numbered 1 to 8 .", + "Output Program": [ + "n0 = 9.0\nn1 = 9.0\nn2 = 8.0\nn3 = 1.0\nn4 = 8.0\nt0 = n2**min(2.0, 5)\nt1 = n2 / t0\nanswer = t1 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.015625" + ], + "split": "train" + }, + { + "Input": "from a pack of 52 cards , 1 card is drawn at random . what is the probability that a jack , queen , or king is drawn ?", + "Output Program": [ + "n0 = 52.0\nn1 = 1.0\nt0 = 3.0 * 4.0\nanswer = t0 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.23076923076923078" + ], + "split": "train" + }, + { + "Input": "there are 18 balls in a jar . you take out 3 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 / 5 . how many blue balls were there in the beginning ?", + "Output Program": [ + "n0 = 18.0\nn1 = 3.0\nn2 = 1.0\nn3 = 5.0\nt0 = n0 - n1\nt1 = t0 / n3\nanswer = n1 + t1\nprint(answer)" + ], + "Output Answer": [ + "6" + ], + "split": "train" + }, + { + "Input": "the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability w that the next couple the maitre ' d seats will not order dessert ?", + "Output Program": [ + "n0 = 60.0\nn1 = 20.0\nt0 = 2.0 * 5.0\nt1 = t0 * t0\nt2 = n0 / t1\nt3 = n1 / t1\nt4 = 1.0 - t3\nt5 = t2 / t4\nt6 = 1.0 - t5\nanswer = t1 * t6\nprint(answer)" + ], + "Output Answer": [ + "25.00000000000001" + ], + "split": "train" + }, + { + "Input": "an urn contains 5 red , 6 blue and 8 green balls . 3 balls are randomly selected from the urn , find the probability that the drawn ball are 2 blue and 1 red ?", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 6.0\nn2 = 8.0\nn3 = 3.0\nn4 = 2.0\nn5 = 1.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n1, n4)\nt2 = scipy.special.comb(n0, n5)\nt3 = n2 + t0\nt4 = t1 * t2\nt5 = scipy.special.comb(t3, n3)\nanswer = t4 / t5\nprint(answer)" + ], + "Output Answer": [ + "0.07739938080495357" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 6 . for a group of 200 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 6.0\nn3 = 200.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "115.74074074074076" + ], + "split": "train" + }, + { + "Input": "how many different ways can 2 students be seated in a row of 6 desks , so that there is always at least one empty desk between the students ?", + "Output Program": [ + "import math\nn0 = 2.0\nn1 = 6.0\nt0 = n1 - 1.0\nanswer = math.factorial(int(max(t0, n0))) / math.factorial(int(max(t0, n0) - min(t0, n0))) # find all permutations\nprint(answer)" + ], + "Output Answer": [ + "20" + ], + "split": "train" + }, + { + "Input": "if 2 cards are selected at random from the deck of 52 cards then what is the probability of one of the selected cards will be king and other will be 10 ? a deck of cards has a total of 52 cards , consisting of 4 suits ; ( spades ( black ) , hearts ( red ) , diamond ( red ) s , and clubs ( black ) ) ; and 13 cards including 1 king , 1 queen and 1 jack in each suit", + "Output Program": [ + "n0 = 2.0\nn1 = 52.0\nn2 = 10.0\nn3 = 52.0\nn4 = 4.0\nn5 = 13.0\nn6 = 1.0\nn7 = 1.0\nn8 = 1.0\nt0 = n4 - 1.0\nt1 = n1 - 1.0\nt2 = t0 * t0\nt3 = n1 * t1\nt4 = t2 * 2.0\nanswer = t4 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.006787330316742082" + ], + "split": "train" + }, + { + "Input": "when 3 fair coins are tossed together , what is the probability of getting 2 tails ?", + "Output Program": [ + "n0 = 3.0\nn1 = 2.0\nt0 = 2.0**min(n0, 5)\nanswer = 4.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.5" + ], + "split": "train" + }, + { + "Input": "a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability x that out of the two , one is white and one is blue ?", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n0 + n1\nt1 = n1 + 2.0\nt2 = n0 * n1\nt3 = t0 * t1\nt4 = t3 / 2.0\nanswer = t2 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.5357142857142857" + ], + "split": "train" + }, + { + "Input": "in a throw of a coin find the probability of getting a tail ?", + "Output Program": [ + "\nanswer = 1.0 / 2.0\nprint(answer)" + ], + "Output Answer": [ + "0.5" + ], + "split": "train" + }, + { + "Input": "a certain club has 10 members , including parry . one of the 10 members is chosen at random to be the president , one of the remaining 9 members is to be chosen at random to be the secretary , and one of the remaining 8 members is to be chosen at random to be the treasurer . what is the probability that parry will be either the member chose to be secretary or the member chose to be treasurer ?", + "Output Program": [ + "n0 = 10.0\nn1 = 10.0\nn2 = 9.0\nn3 = 8.0\nt0 = 1.0 / n0\nanswer = t0 + t0\nprint(answer)" + ], + "Output Answer": [ + "0.2" + ], + "split": "train" + }, + { + "Input": "kim has 8 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 16 shoes , what is the probability that she will select 2 shoes of the same color ?", + "Output Program": [ + "import scipy\nn0 = 8.0\nn1 = 2.0\nn2 = 16.0\nn3 = 2.0\nt0 = scipy.special.comb(n2, n1)\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.06666666666666667" + ], + "split": "train" + }, + { + "Input": "kim has 9 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 18 shoes , what is the probability that she will select 2 shoes of the same color ?", + "Output Program": [ + "import scipy\nn0 = 9.0\nn1 = 2.0\nn2 = 18.0\nn3 = 2.0\nt0 = scipy.special.comb(n2, n1)\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.058823529411764705" + ], + "split": "train" + }, + { + "Input": "a bag contains 2 white marbles and 2 black marbles . if each of 2 girls and 2 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ?", + "Output Program": [ + "import scipy\nn0 = 2.0\nn1 = 2.0\nn2 = 2.0\nn3 = 2.0\nt0 = n0 + n0\nt1 = scipy.special.comb(t0, n0)\nanswer = 2.0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "the main line train starts at 5.00 am and the harbor line train starts at 5.02 am . each train has the frequency of 10 minutes . if a guy goes in the morning at a random time what is the probability of he getting main line train ?", + "Output Program": [ + "n0 = 5.0\nn1 = 5.02\nn2 = 10.0\nt0 = n2 - 2.0\nanswer = t0 / n2\nprint(answer)" + ], + "Output Answer": [ + "0.8" + ], + "split": "train" + }, + { + "Input": "kim has 7 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 14 shoes , what is the probability that she will select 2 shoes of the same color ?", + "Output Program": [ + "import scipy\nn0 = 7.0\nn1 = 2.0\nn2 = 14.0\nn3 = 2.0\nt0 = scipy.special.comb(n2, n1)\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.07692307692307693" + ], + "split": "train" + }, + { + "Input": "there are 7 books on a shelf , of which 2 are paperbacks and 5 are hardbacks . how many possible selections of 4 books from this shelf include at least one paperback ?", + "Output Program": [ + "import scipy\nn0 = 7.0\nn1 = 2.0\nn2 = 5.0\nn3 = 4.0\nt0 = scipy.special.comb(n0, n3)\nt1 = scipy.special.comb(n2, n3)\nanswer = t0 - t1\nprint(answer)" + ], + "Output Answer": [ + "30" + ], + "split": "train" + }, + { + "Input": "there are 4 red shoes & 6 green shoes . if two of red shoes are drawn what is the probability of getting red shoes", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 6.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(t0, 2.0)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.13333333333333333" + ], + "split": "train" + }, + { + "Input": "from a group of 5 boys and 5 girls , 6 children are to be randomly selected . what is the probability that 3 boys and 3 girls will be selected ?", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 5.0\nn2 = 6.0\nn3 = 3.0\nn4 = 3.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(n0, 2.0)\nt3 = scipy.special.comb(t0, n2)\nt4 = t1 * t2\nanswer = t4 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.47619047619047616" + ], + "split": "train" + }, + { + "Input": "a box contains 4 red balls and 4 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ?", + "Output Program": [ + "n0 = 4.0\nn1 = 4.0\nt0 = n0 + n1\nanswer = t0 / n1\nprint(answer)" + ], + "Output Answer": [ + "2" + ], + "split": "train" + }, + { + "Input": "a jar contains 3 black , 3 white and 1 green balls . if you pick two balls at the same time , what ' s the probability that one ball is black and one is white ?", + "Output Program": [ + "n0 = 3.0\nn1 = 3.0\nn2 = 1.0\nt0 = n0 + n0\nt1 = n2 + t0\nt2 = t0 / t1\nt3 = t1 - n2\nt4 = n0 / t3\nanswer = t4 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.42857142857142855" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 8 . for a group of 200 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 8.0\nn3 = 200.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "133.984375" + ], + "split": "train" + }, + { + "Input": "in 15 year \u2019 s time the combined age of my 4 brothers will be 107 . what will it be in 6 year \u2019 s time ?", + "Output Program": [ + "n0 = 15.0\nn1 = 4.0\nn2 = 107.0\nn3 = 6.0\nt0 = n0 - n3\nt1 = n1 * t0\nanswer = n2 - t1\nprint(answer)" + ], + "Output Answer": [ + "71" + ], + "split": "train" + }, + { + "Input": "a , b , c , d , e sitting in a row what is the probability that a & b sitting adjacent .", + "Output Program": [ + "import math\nt0 = 2.0 * 4.0\nt1 = math.factorial(int(max(3.0, 3.0))) / math.factorial(int(max(3.0, 3.0) - min(3.0, 3.0))) # find all permutations\nt2 = math.factorial(int(max(5.0, 5.0))) / math.factorial(int(max(5.0, 5.0) - min(5.0, 5.0))) # find all permutations\nt3 = t0 * t1\nanswer = t3 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.4" + ], + "split": "train" + }, + { + "Input": "on a ranch , a rancher can place a loop of rope , called a lasso , once in every 2 throws around a cow \u2019 s neck . what is the probability that the rancher will be able to place a lasso around a cow \u2019 s neck at least once in 3 attempts ?", + "Output Program": [ + "n0 = 2.0\nn1 = 3.0\nt0 = 1.0 / n0\nt1 = t0**min(n1, 5)\nanswer = 1.0 - t1\nprint(answer)" + ], + "Output Answer": [ + "0.875" + ], + "split": "train" + }, + { + "Input": "in how many ways 3 boys and 3 girls can be seated in a row so that they are alternate .", + "Output Program": [ + "import math\nn0 = 3.0\nn1 = 3.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "36" + ], + "split": "train" + }, + { + "Input": "three unbiased coins are tossed . what is the probability of getting at most two heads ?", + "Output Program": [ + "t0 = 2.0**min(3.0, 5)\nt1 = 1.0 / t0\nanswer = 1 - t1\nprint(answer)" + ], + "Output Answer": [ + "0.875" + ], + "split": "train" + }, + { + "Input": "if x is to be chosen at random from the set { 1 , 2 , 3 , 4 } and y is to be chosen at random from the set { 5 , 6 } , what is the probability that xy will be even ?", + "Output Program": [ + "import scipy\nn0 = 1.0\nn1 = 2.0\nn2 = 3.0\nn3 = 4.0\nn4 = 5.0\nn5 = 6.0\nt0 = scipy.special.comb(n3, n1)\nt1 = scipy.special.comb(n1, n0)\nt2 = n1**min(n1, 5)\nt3 = t0 * t1\nanswer = t2 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "what is the probability q of randomly selecting an arrangement of the letters of the wordmediterraneanin which the first letter is e and the last letter is r ?", + "Output Program": [ + "import math\nt0 = math.factorial(min(15, int(2.0)))\nt1 = math.factorial(min(15, int(3.0)))\nt2 = 3.0 * 4.0\nt3 = t0**min(3.0, 5)\nt4 = t2 - 1.0\nt5 = t4 + 2.0\nt6 = math.factorial(min(15, int(t4)))\nt7 = t1 * t3\nt8 = t6 / t3\nt9 = math.factorial(min(15, int(t5)))\nt10 = t9 / t7\nanswer = t8 / t10\nprint(answer)" + ], + "Output Answer": [ + "0.038461538461538464" + ], + "split": "train" + }, + { + "Input": "there are 4 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 4.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(t0, 2.0)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.21428571428571427" + ], + "split": "train" + }, + { + "Input": "a bag contains 3 red , 4 black and 2 white balls . what is the probability of drawing a red and a white ball in two successive draws , each ball being put back after it is drawn ?", + "Output Program": [ + "n0 = 3.0\nn1 = 4.0\nn2 = 2.0\nt0 = n0 + n1\nt1 = n2 + t0\nt2 = n0 / t1\nt3 = n2 / t1\nt4 = t2 * t3\nanswer = t4 * 2.0\nprint(answer)" + ], + "Output Answer": [ + "0.14814814814814814" + ], + "split": "train" + }, + { + "Input": "if a coin is flipped , the probability that the coin will land tails is 1 / 2 . if the coin is flipped 3 times , what is the probability that it will land tails up on the first flip and not on the last 2 flips ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 3.0\nn3 = 2.0\nt0 = n1**min(3.0, 5)\nanswer = 1 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.125" + ], + "split": "train" + }, + { + "Input": "a box contains 3 red balls and 4 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ?", + "Output Program": [ + "n0 = 3.0\nn1 = 4.0\nt0 = n0 + n1\nanswer = t0 / 4.0\nprint(answer)" + ], + "Output Answer": [ + "1.75" + ], + "split": "train" + }, + { + "Input": "a bag consists of 20 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ?", + "Output Program": [ + "n0 = 20.0\nn1 = 5.0\nn2 = 9.0\nt0 = n1 + n2\nt1 = n0 - t0\nt2 = n2 + t1\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.75" + ], + "split": "train" + }, + { + "Input": "a question paper has 2 parts , a & b , each containing 10 questions . if a student has to choose 8 from part a & 5 from part b , in how many ways can he choose the questions ?", + "Output Program": [ + "import scipy\nn0 = 2.0\nn1 = 10.0\nn2 = 8.0\nn3 = 5.0\nt0 = scipy.special.comb(n1, n2)\nt1 = scipy.special.comb(n1, n3)\nt2 = t0 * t1\nanswer = t2 / n1\nprint(answer)" + ], + "Output Answer": [ + "1134" + ], + "split": "train" + }, + { + "Input": "a meeting has to be conducted with 4 managers . find the number of ways in which the managers may be selected from among 7 managers , if there are 2 managers who refuse to attend the meeting together .", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 7.0\nn2 = 2.0\nt0 = scipy.special.comb(n1, n0)\nt1 = n1 - n2\nt2 = scipy.special.comb(t1, n2)\nanswer = t0 - t2\nprint(answer)" + ], + "Output Answer": [ + "25" + ], + "split": "train" + }, + { + "Input": "in how many ways 4 boys and 1 girls can be seated in a row so that they are alternate .", + "Output Program": [ + "import math\nn0 = 4.0\nn1 = 1.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "24" + ], + "split": "train" + }, + { + "Input": "in how many ways 4 boys and 5 girls can be seated in a row so that they are alternate .", + "Output Program": [ + "import math\nn0 = 4.0\nn1 = 5.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "2880" + ], + "split": "train" + }, + { + "Input": "an unbaised coin is tossed until it shows up the same face in 2 consicative throws wat is the probability that the no of tosses is not more than 4 .", + "Output Program": [ + "n0 = 2.0\nn1 = 4.0\nt0 = n0 * n1\nt1 = 1.0 / t0\nanswer = 1 - t1\nprint(answer)" + ], + "Output Answer": [ + "0.875" + ], + "split": "train" + }, + { + "Input": "in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what is theprobability that the extracted ball would turn onto be white , if all possible hypothesisconcerning the color of the balls that initially in the bag were equally possible ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nt0 = n0 + n1\nt1 = n1 / 2.0\nt2 = n1 / t0\nt3 = n0 / t0\nt4 = t2 + t3\nt5 = t4 + t1\nt6 = n1 + t5\nanswer = t6 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.625" + ], + "split": "train" + }, + { + "Input": "from a group of 3 women and 3 girls , 4 are to be randomly selected . what is the probability that equal numbers of women and girls will be selected ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 3.0\nn2 = 4.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(t0, n2)\nt3 = t1 * t1\nanswer = t3 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.6" + ], + "split": "train" + }, + { + "Input": "a bag consists of 60 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ?", + "Output Program": [ + "n0 = 60.0\nn1 = 5.0\nn2 = 9.0\nt0 = n1 + n2\nt1 = n0 - t0\nt2 = n2 + t1\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.9166666666666666" + ], + "split": "train" + }, + { + "Input": "a meeting has to be conducted with 4 managers . find the number of ways in which the managers may be selected from among 8 managers , if there are 2 managers who refuse to attend the meeting together .", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 8.0\nn2 = 2.0\nt0 = scipy.special.comb(n1, n0)\nt1 = n1 - n2\nt2 = scipy.special.comb(t1, n2)\nanswer = t0 - t2\nprint(answer)" + ], + "Output Answer": [ + "55" + ], + "split": "train" + }, + { + "Input": "when tossed , a certain coin has an equal probability of landing on either side . if the coin is tossed 4 times , what is the probability that it will land on the same side each time ?", + "Output Program": [ + "n0 = 4.0\nt0 = 1.0 / 2.0\nt1 = t0**min(n0, 5)\nanswer = t1 + t1\nprint(answer)" + ], + "Output Answer": [ + "0.125" + ], + "split": "train" + }, + { + "Input": "a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability q that all 3 jellybeans removed from the bag are blue ?", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 10.0\nn2 = 3.0\nn3 = 3.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, n2)\nt2 = scipy.special.comb(t0, n2)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.10526315789473684" + ], + "split": "train" + }, + { + "Input": "find the probability that a number selected from numbers 1 , 2 , 3 , . . . , 50 is a prime number , when each of the given numbers is equally likely to be selected ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 3.0\nn3 = 50.0\nt0 = 3.0 * 5.0\nanswer = t0 / n3\nprint(answer)" + ], + "Output Answer": [ + "0.3" + ], + "split": "train" + }, + { + "Input": "on a ranch , a rancher can place a loop of rope , called a lasso , once in every 2 throws around a cow \u2019 s neck . what is the probability that the rancher will be able to place a lasso around a cow \u2019 s neck at least once in 4 attempts ?", + "Output Program": [ + "n0 = 2.0\nn1 = 4.0\nt0 = 1.0 / n0\nt1 = t0**min(n1, 5)\nanswer = 1.0 - t1\nprint(answer)" + ], + "Output Answer": [ + "0.9375" + ], + "split": "train" + }, + { + "Input": "a quiz has one mcq question with a , b and c as options . and two questions with true / false answers . what is the probability of giving all 3 answers correct ?", + "Output Program": [ + "n0 = 3.0\nt0 = 1.0 / 2.0\nt1 = 1.0 / 3.0\nt2 = t1 * t0\nanswer = t0 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.08333333333333333" + ], + "split": "train" + }, + { + "Input": "in how many ways 4 boys and 3 girls can be seated in a row so that they are alternative ?", + "Output Program": [ + "import math\nn0 = 4.0\nn1 = 3.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "144" + ], + "split": "train" + }, + { + "Input": "when you draw 2 dices together , find the probability of getting a total of 6 ?", + "Output Program": [ + "n0 = 2.0\nn1 = 6.0\nt0 = n1**min(n0, 5)\nt1 = n0 / t0\nt2 = 3.0 / t0\nanswer = t1 + t2\nprint(answer)" + ], + "Output Answer": [ + "0.1388888888888889" + ], + "split": "train" + }, + { + "Input": "when tossed , a certain coin has an equal probability of landing on either side . if the coin is tossed 5 times , what is the probability that it will land on the same side each time ?", + "Output Program": [ + "n0 = 5.0\nt0 = 1.0 / 2.0\nt1 = t0**min(n0, 5)\nanswer = t1 + t1\nprint(answer)" + ], + "Output Answer": [ + "0.0625" + ], + "split": "train" + }, + { + "Input": "set a contains 4 different positive odd integers and 6 different positive even integers . set b contains two different positive odd integers and 3 different positive even integers . if one integer from set a and one integer from set b are chosen at random , what is the probability that the product of the chosen integers is even ?", + "Output Program": [ + "n0 = 4.0\nn1 = 6.0\nn2 = 3.0\nt0 = 2.0 + 3.0\nt1 = n0 + n1\nt2 = 2.0 / t0\nt3 = n0 / t1\nt4 = t2 * t3\nanswer = 1.0 - t4\nprint(answer)" + ], + "Output Answer": [ + "0.84" + ], + "split": "train" + }, + { + "Input": "two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability w that jane will be on the same committee as albert ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 3.0\nt0 = scipy.special.comb(4.0, 1.0)\nt1 = scipy.special.comb(n0, n1)\nt2 = 5.0 * 5.0\nt3 = t0 * 2.0\nt4 = t2 * 4.0\nt5 = t3 / t1\nanswer = t5 * t4\nprint(answer)" + ], + "Output Answer": [ + "40" + ], + "split": "train" + }, + { + "Input": "a box contains 3 red balls and 3 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ?", + "Output Program": [ + "n0 = 3.0\nn1 = 3.0\nt0 = n0 + n1\nanswer = t0 / n1\nprint(answer)" + ], + "Output Answer": [ + "2" + ], + "split": "train" + }, + { + "Input": "from a pack of 52 cards , two cards are drawn together at random . what is the probability of one is a spade and one is a heart ?", + "Output Program": [ + "import scipy\nn0 = 52.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = n0 / 4.0\nt2 = t1 * t1\nanswer = t2 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.12745098039215685" + ], + "split": "train" + }, + { + "Input": "bill and jane play a simple game involving two fair dice , each of which has 6 sides numbered from 1 to 6 ( with an equal chance of landing on any side ) . bill rolls the dice and his score is the total of the two dice . jane then rolls the dice and her score is the total of her two dice . if jane \u2019 s score is higher than bill \u2019 s , she wins the game . what is the probability the jane will win the game ?", + "Output Program": [ + "n0 = 6.0\nn1 = 1.0\nn2 = 6.0\nt0 = n0 * n0\nt1 = n0 - n1\nt2 = t1 * 2.0\nt3 = n0 * t1\nt4 = t0**min(2.0, 5)\nt5 = n1 + t2\nt6 = t5 * t3\nt7 = t6 / n0\nt8 = t7 * 2.0\nt9 = t8 + t0\nt10 = t9 / t4\nt11 = n1 - t10\nanswer = t11 / 2.0\nprint(answer)" + ], + "Output Answer": [ + "0.44367283950617287" + ], + "split": "train" + }, + { + "Input": "a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability q that out of the two , one is white and one is blue ?", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n0 + n1\nt1 = n1 + 2.0\nt2 = n0 * n1\nt3 = t0 * t1\nt4 = t3 / 2.0\nanswer = t2 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.5357142857142857" + ], + "split": "train" + }, + { + "Input": "6 animals of a circus has to be placed inside in 6 cages one in each cage . if 4 of the cage are too small for 6 of the animal then find the number of ways of caging the animal .", + "Output Program": [ + "n0 = 6.0\nn1 = 6.0\nn2 = 4.0\nn3 = 6.0\nt0 = n0 * n2\nanswer = t0 * 10.0\nprint(answer)" + ], + "Output Answer": [ + "240" + ], + "split": "train" + }, + { + "Input": "if you throw two dice at a same time . can you find the probability of getting sum as 10 of the two numbers shown ?", + "Output Program": [ + "n0 = 10.0\nt0 = 6.0**min(2.0, 5)\nt1 = 1.0 / t0\nt2 = t1 + t1\nanswer = t2 + t1\nprint(answer)" + ], + "Output Answer": [ + "0.08333333333333333" + ], + "split": "train" + }, + { + "Input": "a box contains 20 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is", + "Output Program": [ + "import scipy\nn0 = 20.0\nn1 = 4.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, 2.0)\nt3 = t2 / t0\nanswer = 1.0 - t3\nprint(answer)" + ], + "Output Answer": [ + "0.368421052631579" + ], + "split": "train" + }, + { + "Input": "in a lottery there are 10 prizes and 25 blanks . a lottery is drawn at random . what is the probability of getting a blank ?", + "Output Program": [ + "n0 = 10.0\nn1 = 25.0\nt0 = n0 + n1\nanswer = n1 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.7142857142857143" + ], + "split": "train" + }, + { + "Input": "two letters from the word khantkar are selected at random . what is the probability that at least one of the selected letters is a consonant ?", + "Output Program": [ + "import scipy\nt0 = 2.0 * 4.0\nt1 = scipy.special.comb(t0, 2.0)\nt2 = 1.0 / t1\nanswer = 1 - t2\nprint(answer)" + ], + "Output Answer": [ + "0.9642857142857143" + ], + "split": "train" + }, + { + "Input": "persons a and b . person a picks a random no . from 1 to 1000 . then person b picks a random no . from 1 to 1000 . what is the probability of b getting no . greater then what a has picked .", + "Output Program": [ + "n0 = 1.0\nn1 = 1000.0\nn2 = 1.0\nn3 = 1000.0\nt0 = n0 / n1\nt1 = 1.0 - t0\nanswer = t1 / 2.0\nprint(answer)" + ], + "Output Answer": [ + "0.4995" + ], + "split": "train" + }, + { + "Input": "a certain club has 10 members , including jarry . one of the 10 members is chosen at random to be the president , one of the remaining 9 members is to be chosen at random to be the secretary , and one of the remaining 8 members is to be chosen at random to be the treasurer . what is the probability that jarry will be either the member chose to be secretary or the member chose to be treasurer ?", + "Output Program": [ + "n0 = 10.0\nn1 = 10.0\nn2 = 9.0\nn3 = 8.0\nt0 = 1.0 / n0\nt1 = n2 / n0\nt2 = n3 / n2\nt3 = n3 - 1.0\nt4 = t3 / n3\nt5 = t1 * t2\nt6 = t4 * t5\nt7 = t0 + t6\nanswer = 1.0 - t7\nprint(answer)" + ], + "Output Answer": [ + "0.20000000000000007" + ], + "split": "train" + }, + { + "Input": "two dice are thrown simultaneously . what is the probability of getting two numbers whose product is not an even ?", + "Output Program": [ + "t0 = 3.0 + 3.0\nt1 = 3.0 / t0\nanswer = t1 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.25" + ], + "split": "train" + }, + { + "Input": "if 4 consecutive letters are selected at random from the english alphabet , then the probability that one of the letters is a vowel is ?", + "Output Program": [ + "n0 = 4.0\nt0 = 26.0 - 3.0\nt1 = 6.0 / t0\nanswer = 1.0 - t1\nprint(answer)" + ], + "Output Answer": [ + "0.7391304347826086" + ], + "split": "train" + }, + { + "Input": "on a windy day , out of every 3 sand dunes formed , only 1 remains . out of 5 blown - out sand - dunes , 1 has treasure and only 2 out of 3 sand dunes formed have lucky coupons . find the probability that the blown - out sand dune will contain both the things .", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 5.0\nn3 = 1.0\nn4 = 2.0\nn5 = 3.0\nt0 = n4 / n0\nt1 = n1 / n2\nt2 = n1 / n0\nt3 = t0 * t1\nt4 = 1.0 - t2\nanswer = t3 * t4\nprint(answer)" + ], + "Output Answer": [ + "0.08888888888888889" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 700 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 10.0\nn3 = 700.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "510.3" + ], + "split": "train" + }, + { + "Input": "in a single throw of a die , what is probability of getting a number greater than 4 ?", + "Output Program": [ + "n0 = 4.0\nt0 = n0 + 2.0\nanswer = 2.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability y that all 3 jellybeans removed from the bag are blue ?", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 10.0\nn2 = 3.0\nn3 = 3.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, n2)\nt2 = scipy.special.comb(t0, n2)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.10526315789473684" + ], + "split": "train" + }, + { + "Input": "the chance of rain on any given day in tel - aviv is 50 % . what is the probability a that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ?", + "Output Program": [ + "import scipy\nn0 = 50.0\nn1 = 4.0\nn2 = 6.0\nt0 = scipy.special.comb(n2, n1)\nt1 = 1.0 / 2.0\nt2 = t1**min(n1, 5)\nt3 = t1**min(2.0, 5)\nt4 = t2 * t3\nt5 = 1 / t4\nanswer = t0 / t5\nprint(answer)" + ], + "Output Answer": [ + "0.234375" + ], + "split": "train" + }, + { + "Input": "how many different ways can 2 students be seated in a row of 7 desks , so that there is always at least one empty desk between the students ?", + "Output Program": [ + "import math\nn0 = 2.0\nn1 = 7.0\nt0 = n1 - 1.0\nanswer = math.factorial(int(max(t0, n0))) / math.factorial(int(max(t0, n0) - min(t0, n0))) # find all permutations\nprint(answer)" + ], + "Output Answer": [ + "30" + ], + "split": "train" + }, + { + "Input": "in 15 year \u2019 s time the combined age of my 4 brothers will be 107 . what will it be in the 6 year \u2019 s time ?", + "Output Program": [ + "n0 = 15.0\nn1 = 4.0\nn2 = 107.0\nn3 = 6.0\nt0 = n1 * n3\nt1 = n0 * n1\nt2 = n2 - t1\nanswer = t0 + t2\nprint(answer)" + ], + "Output Answer": [ + "71" + ], + "split": "train" + }, + { + "Input": "what is the probability that the sum of two dice will yield a 8 , and then when both are thrown again , their sum will again yield a 8 ? assume that each die has 7 sides with faces numbered 1 to 7 .", + "Output Program": [ + "n0 = 8.0\nn1 = 8.0\nn2 = 7.0\nn3 = 1.0\nn4 = 7.0\nt0 = n2**min(2.0, 5)\nt1 = n2 / t0\nanswer = t1 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.02040816326530612" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 250 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 10.0\nn3 = 250.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "182.25" + ], + "split": "train" + }, + { + "Input": "in how many ways can 11 books on english and 9 books on french be placed in a row on a shelf so that two books on french may not be together ?", + "Output Program": [ + "import math\nimport scipy\nn0 = 11.0\nn1 = 9.0\nt0 = n0 - 1.0\nt1 = scipy.special.comb(t0, 3.0)\nt2 = scipy.special.comb(t0, 1.0)\nt3 = math.factorial(int(max(t0, 2.0))) / math.factorial(int(max(t0, 2.0) - min(t0, 2.0))) # find all permutations\nt4 = t1 + t3\nanswer = t4 + t2\nprint(answer)" + ], + "Output Answer": [ + "220" + ], + "split": "train" + }, + { + "Input": "two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability r that jane will be on the same committee as albert ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 3.0\nt0 = scipy.special.comb(4.0, 1.0)\nt1 = scipy.special.comb(n0, n1)\nt2 = 5.0 * 5.0\nt3 = t0 * 2.0\nt4 = t2 * 4.0\nt5 = t3 / t1\nanswer = t5 * t4\nprint(answer)" + ], + "Output Answer": [ + "40" + ], + "split": "train" + }, + { + "Input": "if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first 2 flips but not on the last 3 flips ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 5.0\nn3 = 2.0\nn4 = 3.0\nt0 = n0 / n1\nanswer = t0**min(n2, 5)\nprint(answer)" + ], + "Output Answer": [ + "0.03125" + ], + "split": "train" + }, + { + "Input": "there are 7 red shoes & 3 green shoes . if two of red shoes are drawn what is the probability of getting red shoes", + "Output Program": [ + "import scipy\nn0 = 7.0\nn1 = 3.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(t0, 2.0)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.4666666666666667" + ], + "split": "train" + }, + { + "Input": "the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability q that the next couple the maitre ' d seats will not order dessert ?", + "Output Program": [ + "n0 = 60.0\nn1 = 20.0\nt0 = 2.0 * 5.0\nt1 = t0 * t0\nt2 = n0 / t1\nt3 = n1 / t1\nt4 = 1.0 - t3\nt5 = t2 / t4\nt6 = 1.0 - t5\nanswer = t1 * t6\nprint(answer)" + ], + "Output Answer": [ + "25.00000000000001" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 300 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 10.0\nn3 = 300.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "218.70000000000002" + ], + "split": "train" + }, + { + "Input": "if x is to be chosen at random from the set { 1 , 2 , 3 , 4 } and y is to be chosen at random from the set { 5 , 6 , 7 } , what is the probability that xy will be odd ?", + "Output Program": [ + "import scipy\nn0 = 1.0\nn1 = 2.0\nn2 = 3.0\nn3 = 4.0\nn4 = 5.0\nn5 = 6.0\nn6 = 7.0\nt0 = scipy.special.comb(n3, n1)\nt1 = scipy.special.comb(n1, n0)\nt2 = n1**min(n1, 5)\nt3 = t0 * t1\nanswer = t2 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "a box contains 2 red balls and 5 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ?", + "Output Program": [ + "n0 = 2.0\nn1 = 5.0\nt0 = n0 + n1\nanswer = t0 / n1\nprint(answer)" + ], + "Output Answer": [ + "1.4" + ], + "split": "train" + }, + { + "Input": "there are 8 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed .", + "Output Program": [ + "import scipy\nn0 = 8.0\nn1 = 2.0\nt0 = scipy.special.comb(n0, n1)\nanswer = t0 + t0\nprint(answer)" + ], + "Output Answer": [ + "56" + ], + "split": "train" + }, + { + "Input": "three coins are tossed . find the probability of at least 2 tails ?", + "Output Program": [ + "n0 = 2.0\nt0 = 2.0**min(n0, 5)\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.5" + ], + "split": "train" + }, + { + "Input": "a letter lock consists of 3 rings each marked with 6 different letters . the number of distinct unsuccessful attempts to open the lock is at the most - .", + "Output Program": [ + "n0 = 3.0\nn1 = 6.0\nt0 = n1 * n1\nt1 = n1 * t0\nanswer = t1 - 1.0\nprint(answer)" + ], + "Output Answer": [ + "215" + ], + "split": "train" + }, + { + "Input": "a meeting has to be conducted with 5 managers . find the number of ways in which the managers may be selected from among 7 managers , if there are 2 managers who refuse to attend the meeting together .", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 7.0\nn2 = 2.0\nt0 = scipy.special.comb(n1, n0)\nt1 = n1 - n2\nt2 = scipy.special.comb(t1, n2)\nanswer = t0 - t2\nprint(answer)" + ], + "Output Answer": [ + "11" + ], + "split": "train" + }, + { + "Input": "a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability g that the first ball picked up is odd numbered ?", + "Output Program": [ + "n0 = 100.0\nn1 = 1.0\nn2 = 100.0\nn3 = 3.0\nn4 = 3.0\n\nanswer = 2.0 / n3\nprint(answer)" + ], + "Output Answer": [ + "0.6666666666666666" + ], + "split": "train" + }, + { + "Input": "the probability that a number selected at random from the first 50 natural numbers is a composite number is ?", + "Output Program": [ + "n0 = 50.0\nt0 = 3.0 * 6.0\nt1 = t0 - 1.0\nt2 = t1 * 2.0\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.68" + ], + "split": "train" + }, + { + "Input": "45 girls and 60 boys are present at a party . there are 17 girl - boy are in relationship among them . if a girl and a boy is selected at random , what is the probability that they will be not a couple ?", + "Output Program": [ + "import scipy\nn0 = 45.0\nn1 = 60.0\nn2 = 17.0\nt0 = scipy.special.comb(n0, 1.0)\nt1 = scipy.special.comb(n1, 1.0)\nt2 = scipy.special.comb(n2, 1.0)\nt3 = t0 * t1\nt4 = t3 - t2\nanswer = t4 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.9937037037037038" + ], + "split": "train" + }, + { + "Input": "a store has 10 bottles of juice , including 6 bottles of apple juice . in the evening , 6 bottles of juice are sold one by one . what is the probability of selling 2 bottles of apple juice among the 6 bottles ? assume that every bottle has an equal chance of being bought .", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 6.0\nn2 = 6.0\nn3 = 2.0\nn4 = 6.0\nt0 = scipy.special.comb(n1, n3)\nt1 = scipy.special.comb(n0, n1)\nanswer = t0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.07142857142857142" + ], + "split": "train" + }, + { + "Input": "when two dice are rolled , what is the probability that the sum is either 7 or 11 ?", + "Output Program": [ + "n0 = 7.0\nn1 = 11.0\nt0 = n0 + 1.0\nt1 = 6.0 * 6.0\nanswer = t0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.2222222222222222" + ], + "split": "train" + }, + { + "Input": "two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability q that jane will be on the same committee as albert ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 3.0\nt0 = scipy.special.comb(4.0, 1.0)\nt1 = scipy.special.comb(n0, n1)\nt2 = 5.0 * 5.0\nt3 = t0 * 2.0\nt4 = t2 * 4.0\nt5 = t3 / t1\nanswer = t5 * t4\nprint(answer)" + ], + "Output Answer": [ + "40" + ], + "split": "train" + }, + { + "Input": "if g is to be chosen at random from the set { 5,8 , 7,1 } and k is to be chosen at random from the set { 14 , 8,3 } , what is the probability that gk will be even ?", + "Output Program": [ + "n0 = 5.0\nn1 = 8.0\nn2 = 7.0\nn3 = 1.0\nn4 = 14.0\nn5 = 8.0\nn6 = 3.0\nt0 = 3.0 * 4.0\nanswer = t0 / t0\nprint(answer)" + ], + "Output Answer": [ + "1" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 500 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 10.0\nn3 = 500.0\nn4 = 3.0\nt0 = 1.0 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "364.5" + ], + "split": "train" + }, + { + "Input": "kim has 5 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 10 shoes , what is the probability that she will select 2 shoes of the same color ?", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 2.0\nn2 = 10.0\nn3 = 2.0\nt0 = scipy.special.comb(n2, n1)\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.1111111111111111" + ], + "split": "train" + }, + { + "Input": "find the probability that a leap year selected at random will have 53 mondays", + "Output Program": [ + "n0 = 53.0\nt0 = 3.0 + 4.0\nanswer = 2.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.2857142857142857" + ], + "split": "train" + }, + { + "Input": "a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability l that the first ball picked up is odd numbered ?", + "Output Program": [ + "n0 = 100.0\nn1 = 1.0\nn2 = 100.0\nn3 = 3.0\nn4 = 3.0\n\nanswer = 2.0 / n3\nprint(answer)" + ], + "Output Answer": [ + "0.6666666666666666" + ], + "split": "train" + }, + { + "Input": "two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability e that jane will be on the same committee as albert ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 3.0\nt0 = scipy.special.comb(4.0, 1.0)\nt1 = scipy.special.comb(n0, n1)\nt2 = 5.0 * 5.0\nt3 = t0 * 2.0\nt4 = t2 * 4.0\nt5 = t3 / t1\nanswer = t5 * t4\nprint(answer)" + ], + "Output Answer": [ + "40" + ], + "split": "train" + }, + { + "Input": "there is a 50 % chance jen will visit chile this year , while there is a 50 % chance that she will visit madagascar this year . what is the probability that jen will visit either chile or madagascar this year , but not both ?", + "Output Program": [ + "n0 = 50.0\nn1 = 50.0\nt0 = 2.0 * 5.0\nt1 = t0**min(2.0, 5)\nt2 = n1 / t1\nt3 = n0 / t1\nt4 = t1 - n0\nt5 = t4 / t1\nt6 = t3 * t2\nt7 = t5 * t2\nt8 = t7 + t6\nanswer = t8 * t1\nprint(answer)" + ], + "Output Answer": [ + "50" + ], + "split": "train" + }, + { + "Input": "in how many ways can 21 books on english and 19 books on french be placed in a row on a shelf so that two books on french may not be together ?", + "Output Program": [ + "import math\nimport scipy\nn0 = 21.0\nn1 = 19.0\nt0 = n0 - 1.0\nt1 = scipy.special.comb(t0, 3.0)\nt2 = scipy.special.comb(t0, 1.0)\nt3 = math.factorial(int(max(t0, 2.0))) / math.factorial(int(max(t0, 2.0) - min(t0, 2.0))) # find all permutations\nt4 = t1 + t3\nanswer = t4 + t2\nprint(answer)" + ], + "Output Answer": [ + "1540" + ], + "split": "train" + }, + { + "Input": "john borrowed 3 soccer boots from jake , and forgot them on the field , if peter came across a total of 15 boots on the field jake ' s boots inclusive , and he took 4 boots at random , what is the probability that jake ' s boots were not amongst the 4 taken ? .", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 15.0\nn2 = 4.0\nn3 = 4.0\nt0 = scipy.special.comb(n1, n2)\nt1 = n1 - n0\nt2 = scipy.special.comb(t1, n2)\nanswer = t2 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.3626373626373626" + ], + "split": "train" + }, + { + "Input": "a circular dartboard of radius 1 foot is at a distance of 20 feet from you . you throw a dart at it and it hits the dartboard at some point q in the circle . what is the probability that q is closer to the center of the circle than the periphery ?", + "Output Program": [ + "n0 = 1.0\nn1 = 20.0\nt0 = n0 / 2.0\nt1 = 1.0 * 1.0\nt2 = t0 * t0\nanswer = t2 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.25" + ], + "split": "train" + }, + { + "Input": "in lottery , there are 10 prizes and 25 blanks . a lottery is drawn at random . what is the probability of getting a prize ?", + "Output Program": [ + "n0 = 10.0\nn1 = 25.0\nt0 = n0 + n1\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.2857142857142857" + ], + "split": "train" + }, + { + "Input": "there are 6 red balls and 4 blue balls in a jar . if 3 balls are selected from the jar , what is the probability that all 3 balls selected are blue balls ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 4.0\nn2 = 3.0\nn3 = 3.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n1, n2)\nt2 = scipy.special.comb(t0, n2)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.03333333333333333" + ], + "split": "train" + }, + { + "Input": "the probability that a number selected at random from the first 50 natural numbers is a composite number is - .", + "Output Program": [ + "n0 = 50.0\nt0 = 3.0 * 6.0\nt1 = t0 - 1.0\nt2 = t1 * 2.0\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.68" + ], + "split": "train" + }, + { + "Input": "there are 10 books on a shelf , of which 4 are paperbacks and 6 are hardbacks . how many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback ?", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 4.0\nn2 = 6.0\nn3 = 5.0\nt0 = scipy.special.comb(n0, n3)\nt1 = scipy.special.comb(n2, n3)\nanswer = t0 - t1\nprint(answer)" + ], + "Output Answer": [ + "246" + ], + "split": "train" + }, + { + "Input": "what is the probability that the sum of two dice will yield a 4 , and then when both are thrown again , their sum will again yield a 4 ? assume that each die has 3 sides with faces numbered 1 to 3 .", + "Output Program": [ + "n0 = 4.0\nn1 = 4.0\nn2 = 3.0\nn3 = 1.0\nn4 = 3.0\nt0 = n2**min(2.0, 5)\nt1 = n2 / t0\nanswer = t1 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.1111111111111111" + ], + "split": "train" + }, + { + "Input": "a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability z that the first ball picked up is odd numbered ?", + "Output Program": [ + "n0 = 100.0\nn1 = 1.0\nn2 = 100.0\nn3 = 3.0\nn4 = 3.0\n\nanswer = 2.0 / n3\nprint(answer)" + ], + "Output Answer": [ + "0.6666666666666666" + ], + "split": "train" + }, + { + "Input": "if 4 people are selected from a group of 9 married couples , what is the probability that none of them would be married to each other ?", + "Output Program": [ + "n0 = 4.0\nn1 = 9.0\nt0 = 4.0 * 4.0\nt1 = n1 * 2.0\nt2 = t1 / t1\nt3 = t0 - 2.0\nt4 = t0 - 1.0\nt5 = t1 - 1.0\nt6 = t3 / t0\nt7 = t0 / t5\nt8 = t3 - 2.0\nt9 = t8 / t4\nt10 = t2 * t7\nt11 = t6 * t10\nanswer = t9 * t11\nprint(answer)" + ], + "Output Answer": [ + "0.6588235294117647" + ], + "split": "train" + }, + { + "Input": "from a group of 3 boys and 3 girls , 2 children are to be randomly selected . what is the probability that 1 boy and 1 girl will be selected ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 3.0\nn2 = 2.0\nn3 = 1.0\nn4 = 1.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(n0, 2.0)\nt3 = scipy.special.comb(t0, n2)\nt4 = t1 * t2\nanswer = t4 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.6" + ], + "split": "train" + }, + { + "Input": "tomeka is playing a dice game . if she rolls the same number on her second roll as she rolls on her first , she wins . each roll is with two , fair , 6 - sided dice . if tomeka rolled a 7 on her first roll , what is the probability that she will win on her second roll ?", + "Output Program": [ + "n0 = 6.0\nn1 = 7.0\nt0 = n0 * n0\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.16666666666666666" + ], + "split": "train" + }, + { + "Input": "john and peter are among the 9 players a volleyball coach can choose from to field a 6 - player team . if all 6 players are chosen at random , what is the probability of choosing a team that includes john and peter ?", + "Output Program": [ + "import math\nn0 = 9.0\nn1 = 6.0\nn2 = 6.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nt2 = n0 - 2.0\nt3 = n1 - 2.0\nt4 = n0 - n1\nt5 = math.factorial(min(15, int(t2)))\nt6 = math.factorial(min(15, int(t3)))\nt7 = math.factorial(min(15, int(t4)))\nt8 = t2 - t3\nt9 = math.factorial(min(15, int(t8)))\nt10 = t1 * t7\nt11 = t0 / t10\nt12 = t6 * t9\nt13 = t5 / t12\nanswer = t13 / t11\nprint(answer)" + ], + "Output Answer": [ + "0.4166666666666667" + ], + "split": "train" + }, + { + "Input": "in how many ways 4 boys and 3 girls can be seated in a row so that they are alternate .", + "Output Program": [ + "import math\nn0 = 4.0\nn1 = 3.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "144" + ], + "split": "train" + }, + { + "Input": "in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 3 / 4 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and curved", + "Output Program": [ + "n0 = 3.0\nn1 = 4.0\nn2 = 1.0\nn3 = 2.0\nt0 = n2 / n3\nt1 = n0 / n1\nt2 = 1.0 - t1\nanswer = t0 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.125" + ], + "split": "train" + }, + { + "Input": "a player holds 13 cards of 4 suits , of which 7 are black and 6 are red . there are twice as many diamonds as spades and twice as many hearts as diamonds . how many clubs does he hold", + "Output Program": [ + "n0 = 13.0\nn1 = 4.0\nn2 = 7.0\nn3 = 6.0\nt0 = n1 + 2.0\nt1 = n0 - n2\nt2 = t1 / t0\nanswer = n2 - t2\nprint(answer)" + ], + "Output Answer": [ + "6" + ], + "split": "train" + }, + { + "Input": "a bag consists of 50 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ?", + "Output Program": [ + "n0 = 50.0\nn1 = 5.0\nn2 = 9.0\nt0 = n1 + n2\nt1 = n0 - t0\nt2 = n2 + t1\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.9" + ], + "split": "train" + }, + { + "Input": "a box contains 23 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is", + "Output Program": [ + "import scipy\nn0 = 23.0\nn1 = 4.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, 2.0)\nt3 = t2 / t0\nanswer = 1.0 - t3\nprint(answer)" + ], + "Output Answer": [ + "0.3241106719367589" + ], + "split": "train" + }, + { + "Input": "what is the probability that a number selected from number 1 to 30 is prime number , when each of the given numbers is equally likely to be selected ?", + "Output Program": [ + "n0 = 1.0\nn1 = 30.0\nt0 = 2.0 * 5.0\nanswer = t0 / n1\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "there are 8 students . 4 of them are men and 4 of them are women . if 4 students are selected from the 8 students . what is the probability that the number of men is equal to that of women ?", + "Output Program": [ + "import scipy\nn0 = 8.0\nn1 = 4.0\nn2 = 4.0\nn3 = 4.0\nn4 = 8.0\nt0 = scipy.special.comb(n0, n1)\nt1 = n1 / 2.0\nt2 = scipy.special.comb(n1, t1)\nt3 = t2 * t2\nanswer = t3 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.5142857142857142" + ], + "split": "train" + }, + { + "Input": "a classroom has 3 electric lamps . from a collocation of 10 electric bulbs of which 6 are good , 3 are selected at random and put in the lamps . find the probability that the room is lighted .", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 10.0\nn2 = 6.0\nn3 = 3.0\nt0 = scipy.special.comb(n1, n0)\nt1 = n1 - n2\nt2 = scipy.special.comb(t1, n0)\nt3 = t0 - t2\nanswer = t3 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.9666666666666667" + ], + "split": "train" + }, + { + "Input": "in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what is the probability that the extracted ball would turn on to be white , if all possible hypothesis concerning the color of theballs that initiallyin the bag were equally possible ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nt0 = n0 + n1\nt1 = n1 / 2.0\nt2 = n1 / t0\nt3 = n0 / t0\nt4 = t2 + t3\nt5 = t4 + t1\nt6 = n1 + t5\nanswer = t6 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.625" + ], + "split": "train" + }, + { + "Input": "rhoda tosses a coin 5 times . find the probability of getting exactly 4 heads .", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 4.0\nt0 = scipy.special.comb(n0, n1)\nt1 = 2.0**min(n0, 5)\nanswer = t0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.15625" + ], + "split": "train" + }, + { + "Input": "a card is drawn from a pack of 52 cards . the probability of not getting a face card ?", + "Output Program": [ + "n0 = 52.0\nt0 = 4.0 * 4.0\nt1 = n0 - t0\nanswer = t1 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.6923076923076923" + ], + "split": "train" + }, + { + "Input": "potato chips are on sale , buy 2 get 1 free . there are only 9 bags of chips left ( 4 plain and 5 bbq ) . a shopper randomly grabs 3 bags . what is the probability the shopper will grab exactly 3 bags of bbq chips ?", + "Output Program": [ + "n0 = 2.0\nn1 = 1.0\nn2 = 9.0\nn3 = 4.0\nn4 = 5.0\nn5 = 3.0\nn6 = 3.0\nt0 = n4 / n2\nt1 = n4 - 1.0\nt2 = n2 - 1.0\nt3 = t1 / t2\nt4 = t1 - 1.0\nt5 = t2 - 1.0\nt6 = t4 / t5\nt7 = t0 * t3\nanswer = t6 * t7\nprint(answer)" + ], + "Output Answer": [ + "0.11904761904761904" + ], + "split": "train" + }, + { + "Input": "there are two bags a and b . a contains 6 red flowers and 3 pink flowers . where as bag b contains 2 red flowers and 7 pink flowers . one flower is chosen from a bag randomly . what is the probability that the flower chosen is pink ?", + "Output Program": [ + "n0 = 6.0\nn1 = 3.0\nn2 = 2.0\nn3 = 7.0\nt0 = n0 + n1\nt1 = n2 + n3\nt2 = 1.0 / 2.0\nt3 = 3.0 / t0\nt4 = n3 / t1\nt5 = t3 + t4\nanswer = t5 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.5555555555555556" + ], + "split": "train" + }, + { + "Input": "find the probability that a leap year chosen at random will have 53 sundays", + "Output Program": [ + "n0 = 53.0\nt0 = 3.0 + 4.0\nanswer = 2.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.2857142857142857" + ], + "split": "train" + }, + { + "Input": "if you keep rolling a pair of dice together till a sum of 5 or 7 is obtained , then what is the probability that a sum of 5 comes before a sum of 7 ?", + "Output Program": [ + "n0 = 5.0\nn1 = 7.0\nn2 = 5.0\nn3 = 7.0\nt0 = 4.0 + 6.0\nanswer = 4.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.4" + ], + "split": "train" + }, + { + "Input": "the chance of rain on any given day in tel - aviv is 50 % . what is the probability q that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ?", + "Output Program": [ + "import scipy\nn0 = 50.0\nn1 = 4.0\nn2 = 6.0\nt0 = scipy.special.comb(n2, n1)\nt1 = 1.0 / 2.0\nt2 = t1**min(n1, 5)\nt3 = t1**min(2.0, 5)\nt4 = t2 * t3\nt5 = 1 / t4\nanswer = t0 / t5\nprint(answer)" + ], + "Output Answer": [ + "0.234375" + ], + "split": "train" + }, + { + "Input": "the chance of rain on any given day in tel - aviv is 50 % . what is the probability e that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ?", + "Output Program": [ + "import scipy\nn0 = 50.0\nn1 = 4.0\nn2 = 6.0\nt0 = scipy.special.comb(n2, n1)\nt1 = 1.0 / 2.0\nt2 = t1**min(n1, 5)\nt3 = t1**min(2.0, 5)\nt4 = t2 * t3\nt5 = 1 / t4\nanswer = t0 / t5\nprint(answer)" + ], + "Output Answer": [ + "0.234375" + ], + "split": "train" + }, + { + "Input": "there are 6 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 4.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(t0, 2.0)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "what is the probability that carol and bernie will get selected for a job they both applied for in a company , given that their chances of getting selected is 4 / 5 and 3 / 5 respectively ?", + "Output Program": [ + "n0 = 4.0\nn1 = 5.0\nn2 = 3.0\nn3 = 5.0\nt0 = n0 / n1\nt1 = n2 / n1\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.48" + ], + "split": "train" + }, + { + "Input": "if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is ?", + "Output Program": [ + "t0 = 3.0 / 52.0\nt1 = 52.0 / 4.0\nt2 = t1 / 52.0\nanswer = t0 + t2\nprint(answer)" + ], + "Output Answer": [ + "0.3076923076923077" + ], + "split": "train" + }, + { + "Input": "what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular 9 - sided polygon ) ?", + "Output Program": [ + "n0 = 9.0\nt0 = n0 - 3.0\nt1 = n0 * t0\nt2 = t1 / 2.0\nanswer = n0 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability z that out of the two , one is white and one is blue ?", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n0 + n1\nt1 = n1 + 2.0\nt2 = n0 * n1\nt3 = t0 * t1\nt4 = t3 / 2.0\nanswer = t2 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.5357142857142857" + ], + "split": "train" + }, + { + "Input": "when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 5 times , what is the probability that it will land on either side all 5 times ?", + "Output Program": [ + "n0 = 5.0\nn1 = 5.0\nt0 = 2.0**min(4.0, 5)\nanswer = 1.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.0625" + ], + "split": "train" + }, + { + "Input": "there are 10 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed .", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 2.0\nt0 = scipy.special.comb(n0, n1)\nanswer = t0 + t0\nprint(answer)" + ], + "Output Answer": [ + "90" + ], + "split": "train" + }, + { + "Input": "in how many ways can 21 books on english and 19 books on hindi be placed in a row on a shelf so that two books on hindi may not be together ?", + "Output Program": [ + "import math\nimport scipy\nn0 = 21.0\nn1 = 19.0\nt0 = n0 - 1.0\nt1 = scipy.special.comb(t0, 3.0)\nt2 = scipy.special.comb(t0, 1.0)\nt3 = math.factorial(int(max(t0, 2.0))) / math.factorial(int(max(t0, 2.0) - min(t0, 2.0))) # find all permutations\nt4 = t1 + t3\nanswer = t4 + t2\nprint(answer)" + ], + "Output Answer": [ + "1540" + ], + "split": "train" + }, + { + "Input": "a bag contains 5 white marbles and 5 black marbles . if each of 5 girls and 5 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ?", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 5.0\nn2 = 5.0\nn3 = 5.0\nt0 = n0 + n0\nt1 = scipy.special.comb(t0, n0)\nanswer = 2.0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.007936507936507936" + ], + "split": "train" + }, + { + "Input": "what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular 7 - sided polygon ) ?", + "Output Program": [ + "n0 = 7.0\nt0 = n0 - 3.0\nt1 = n0 * t0\nt2 = t1 / 2.0\nanswer = n0 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.5" + ], + "split": "train" + }, + { + "Input": "what is the probability that a two digit number selected at random will be a multiple of ' 3 ' and not a multiple of ' 5 ' ?", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n1 * 2.0\nt1 = n0 * 2.0\nt2 = n0 * n0\nt3 = n0 * t0\nt4 = t0 * t2\nt5 = t3 - t1\nanswer = t5 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.26666666666666666" + ], + "split": "train" + }, + { + "Input": "what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular 11 - sided polygon ) ?", + "Output Program": [ + "n0 = 11.0\nt0 = n0 - 3.0\nt1 = n0 * t0\nt2 = t1 / 2.0\nanswer = n0 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.25" + ], + "split": "train" + }, + { + "Input": "in a throw of a coin probability of getting a head .", + "Output Program": [ + "\nanswer = 1.0 / 2.0\nprint(answer)" + ], + "Output Answer": [ + "0.5" + ], + "split": "train" + }, + { + "Input": "find the probability that a leap year selected at random will have 53 sundays", + "Output Program": [ + "n0 = 53.0\nt0 = 3.0 + 4.0\nanswer = 2.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.2857142857142857" + ], + "split": "train" + }, + { + "Input": "what is the probability that company g ' s advertisement will be one of the first two be shown during the first commercial break ?", + "Output Program": [ + "t0 = 3.0 * 5.0\nanswer = 2.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.13333333333333333" + ], + "split": "train" + }, + { + "Input": "the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability y that the next couple the maitre ' d seats will not order dessert ?", + "Output Program": [ + "n0 = 60.0\nn1 = 20.0\nt0 = 2.0 * 5.0\nt1 = t0 * t0\nt2 = n0 / t1\nt3 = n1 / t1\nt4 = 1.0 - t3\nt5 = t2 / t4\nt6 = 1.0 - t5\nanswer = t1 * t6\nprint(answer)" + ], + "Output Answer": [ + "25.00000000000001" + ], + "split": "train" + }, + { + "Input": "what is the probability that the sum of two dice will yield a 7 , and then when both are thrown again , their sum will again yield a 7 ? assume that each die has 6 sides with faces numbered 1 to 6 .", + "Output Program": [ + "n0 = 7.0\nn1 = 7.0\nn2 = 6.0\nn3 = 1.0\nn4 = 6.0\nt0 = n2**min(2.0, 5)\nt1 = n2 / t0\nanswer = t1 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.027777777777777776" + ], + "split": "train" + }, + { + "Input": "5 letters have to be put in 5 different envelopes numbered 1 through 5 such that each of the letters go into only 1 envelope . the letter is said to be put in correct position if for example letter 1 goes into envelope 1 . now what is the probability that all letters be put into wrong envelopes ?", + "Output Program": [ + "import math\nn0 = 5.0\nn1 = 5.0\nn2 = 1.0\nn3 = 5.0\nn4 = 1.0\nn5 = 1.0\nn6 = 1.0\nt0 = math.factorial(min(15, int(4.0)))\nt1 = math.factorial(min(15, int(2.0)))\nt2 = math.factorial(min(15, int(3.0)))\nt3 = math.factorial(min(15, int(n0)))\nt4 = 1 / t0\nt5 = 1 / t1\nt6 = 1 / t2\nt7 = 1 / t3\nt8 = t5 - t6\nt9 = t4 + t8\nanswer = t9 - t7\nprint(answer)" + ], + "Output Answer": [ + "0.3666666666666667" + ], + "split": "train" + }, + { + "Input": "in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 2 / 3 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and straight", + "Output Program": [ + "n0 = 2.0\nn1 = 3.0\nn2 = 1.0\nn3 = 2.0\nt0 = n2 / n3\nt1 = n0 / n1\nt2 = n2 - t1\nanswer = t0 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.16666666666666669" + ], + "split": "train" + }, + { + "Input": "in a throw of a coin find the probability of getting a head ?", + "Output Program": [ + "\nanswer = 1.0 / 2.0\nprint(answer)" + ], + "Output Answer": [ + "0.5" + ], + "split": "train" + }, + { + "Input": "the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability r that the next couple the maitre ' d seats will not order dessert ?", + "Output Program": [ + "n0 = 60.0\nn1 = 20.0\nt0 = 2.0 * 5.0\nt1 = t0 * t0\nt2 = n0 / t1\nt3 = n1 / t1\nt4 = 1.0 - t3\nt5 = t2 / t4\nt6 = 1.0 - t5\nanswer = t1 * t6\nprint(answer)" + ], + "Output Answer": [ + "25.00000000000001" + ], + "split": "train" + }, + { + "Input": "if a coin is flipped , the probability that the coin will land tails is 1 / 2 . if the coin is flipped 4 times , what is the probability that it will land tails up on the first 2 flips and not on the last 2 flips ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 4.0\nn3 = 2.0\nn4 = 2.0\nt0 = n1**min(n2, 5)\nanswer = 1 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.0625" + ], + "split": "train" + }, + { + "Input": "kim has 6 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 12 shoes , what is the probability that she will select 2 shoes of the same color ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 2.0\nn2 = 12.0\nn3 = 2.0\nt0 = scipy.special.comb(n2, n1)\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.09090909090909091" + ], + "split": "train" + }, + { + "Input": "in a throw of dice what is the probability of ge \u00e6 \u00ab ing number greater than 2", + "Output Program": [ + "n0 = 2.0\nt0 = 6.0 - n0\nanswer = t0 / 6.0\nprint(answer)" + ], + "Output Answer": [ + "0.6666666666666666" + ], + "split": "train" + }, + { + "Input": "kim has 4 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 8 shoes , what is the probability that she will select 2 shoes of the same color ?", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 2.0\nn2 = 8.0\nn3 = 2.0\nt0 = scipy.special.comb(n2, n1)\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.14285714285714285" + ], + "split": "train" + }, + { + "Input": "there are 11 boys and 10 girls in a class . if three students are selected at random , in how many ways that 3 girl & 2 boys are selected ?", + "Output Program": [ + "import scipy\nn0 = 11.0\nn1 = 10.0\nn2 = 3.0\nn3 = 2.0\nt0 = scipy.special.comb(n0, n3)\nt1 = scipy.special.comb(n1, n2)\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "6600" + ], + "split": "train" + }, + { + "Input": "there are 13 boys and 10 girls in a class . if three students are selected at random , in how many ways that 1 girl & 2 boys are selected ?", + "Output Program": [ + "import scipy\nn0 = 13.0\nn1 = 10.0\nn2 = 1.0\nn3 = 2.0\nt0 = scipy.special.comb(n0, n3)\nt1 = scipy.special.comb(n1, n2)\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "780" + ], + "split": "train" + }, + { + "Input": "how many different pairs of numbers ( s , t ) such that s = 3 t can be obtained if s and t are selected from the set of number { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 }", + "Output Program": [ + "n0 = 3.0\nn1 = 0.0\nn2 = 1.0\nn3 = 2.0\nn4 = 3.0\nn5 = 4.0\nn6 = 5.0\nn7 = 6.0\nn8 = 7.0\nn9 = 8.0\nn10 = 9.0\nn11 = 10.0\nn12 = 11.0\nn13 = 12.0\nt0 = n13 / n0\nanswer = t0 + 1.0\nprint(answer)" + ], + "Output Answer": [ + "5" + ], + "split": "train" + }, + { + "Input": "three unbiased coins are tossed . what is the probability of getting at most 2 heads ?", + "Output Program": [ + "n0 = 2.0\nt0 = 2.0**min(3.0, 5)\nt1 = 1.0 / t0\nanswer = 1 - t1\nprint(answer)" + ], + "Output Answer": [ + "0.875" + ], + "split": "train" + }, + { + "Input": "a bag consists of 20 marbles , of which 6 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ?", + "Output Program": [ + "n0 = 20.0\nn1 = 6.0\nn2 = 9.0\nt0 = n1 + n2\nt1 = n0 - t0\nt2 = n2 + t1\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.7" + ], + "split": "train" + }, + { + "Input": "a father purchased dress for his 3 daughters . the dresses are of same color but diff size and they are kept in dark room . what is probability that all the 3 will not choose there own dress ?", + "Output Program": [ + "import math\nimport scipy\nn0 = 3.0\nn1 = 3.0\nt0 = scipy.special.comb(2.0, 1.0)\nt1 = math.factorial(min(15, int(3.0)))\nanswer = t0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "a child want to buy 5 new toys . if there are 15 toys that satisfy the child ' s criteria , in how many different ways can the child select the 5 new toys if the order of selection does n ' t matter .", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 15.0\nn2 = 5.0\n\nanswer = scipy.special.comb(n1, n0)\nprint(answer)" + ], + "Output Answer": [ + "3003" + ], + "split": "train" + }, + { + "Input": "a bag contains 3 red , 5 yellow and 4 green balls . 3 balls are drawn randomly . what is the probability that balls drawn contain exactly two green balls ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 5.0\nn2 = 4.0\nn3 = 3.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n2, 2.0)\nt2 = n2 + t0\nt3 = scipy.special.comb(t0, 1.0)\nt4 = scipy.special.comb(t2, n0)\nt5 = t1 * t3\nanswer = t5 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.21818181818181817" + ], + "split": "train" + }, + { + "Input": "the letters o , l and t are rearranged to form the word ' lot ' . find its probability .", + "Output Program": [ + "import math\nt0 = math.factorial(min(15, int(3.0)))\nanswer = 1.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.16666666666666666" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 200 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 10.0\nn3 = 200.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "145.8" + ], + "split": "train" + }, + { + "Input": "a jar contains 5 black and 2 white balls . if you pick two balls at the same time , what ' s the probability that one ball is black and one is white ?", + "Output Program": [ + "n0 = 5.0\nn1 = 2.0\nt0 = n0 + n1\nt1 = n0 / t0\nt2 = t0 - 1.0\nt3 = n1 / t2\nanswer = t1 * t3\nprint(answer)" + ], + "Output Answer": [ + "0.23809523809523808" + ], + "split": "train" + }, + { + "Input": "a jar contains 18 balls . 3 blue balls are removed from the jar and not replaced . now the probability of getting a blue ball is 1 / 5 then how many blue balls jar contains initially ?", + "Output Program": [ + "n0 = 18.0\nn1 = 3.0\nn2 = 1.0\nn3 = 5.0\nt0 = n2 / n3\nt1 = n0 - n1\nt2 = t0 * t1\nanswer = n1 + t2\nprint(answer)" + ], + "Output Answer": [ + "6" + ], + "split": "train" + }, + { + "Input": "in a bag containing 5 balls , a white ball was placed and then 1 ball was taken out at random . what is the probability that the extracted ball would turn on to be white , if all possible hypothesis concerning the color of the balls that initiallyin the bag wereequally possible ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nt0 = n0 + n1\nt1 = n1 / 2.0\nt2 = n1 / t0\nt3 = n0 / t0\nt4 = t2 + t3\nt5 = t4 + t1\nt6 = n1 + t5\nanswer = t6 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.625" + ], + "split": "train" + }, + { + "Input": "in a single throw of a die , what is the probability of getting a number greater than 4 ?", + "Output Program": [ + "n0 = 4.0\nt0 = n0 + 2.0\nanswer = 2.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first flip but not on the last 4 flips ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 5.0\nn3 = 4.0\nt0 = n0 / n1\nanswer = t0**min(n2, 5)\nprint(answer)" + ], + "Output Answer": [ + "0.03125" + ], + "split": "train" + }, + { + "Input": "wink , inc . follows a certain procedure that requires two tasks to be finished independently in order for a job to be done . on any given day , there is a 5 / 8 probability that task 1 will be completed on time , and a 3 / 5 probability that task 2 will be completed on time . on a certain day , what is the probability that task 1 will be completed on time , but task 2 will not ?", + "Output Program": [ + "n0 = 5.0\nn1 = 8.0\nn2 = 1.0\nn3 = 3.0\nn4 = 5.0\nn5 = 2.0\nn6 = 1.0\nn7 = 2.0\nt0 = n0 / n1\nt1 = n3 / n0\nt2 = 1.0 - t1\nanswer = t0 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.25" + ], + "split": "train" + }, + { + "Input": "if a coin is flipped , the probability that the coin will land tails is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land tails up on the first 2 flips and not on the last 3 flips ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 5.0\nn3 = 2.0\nn4 = 3.0\nt0 = n1**min(n2, 5)\nanswer = 1 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.03125" + ], + "split": "train" + }, + { + "Input": "out of 15 students in a class , 7 are from maharashtra , 5 are from karnataka , and 3 are from goa . four students are to be selected at random . what are the chances that at least one is from karnataka ?", + "Output Program": [ + "import math\nn0 = 15.0\nn1 = 7.0\nn2 = 5.0\nn3 = 3.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(4.0)))\nt2 = n0 - 4.0\nt3 = n0 - 5.0\nt4 = math.factorial(min(15, int(t2)))\nt5 = math.factorial(min(15, int(t3)))\nt6 = t3 - 4.0\nt7 = t0 / t4\nt8 = math.factorial(min(15, int(t6)))\nt9 = t7 / t1\nt10 = t5 / t8\nt11 = t10 / t1\nt12 = t9 - t11\nanswer = t12 / t9\nprint(answer)" + ], + "Output Answer": [ + "0.8461538461538461" + ], + "split": "train" + }, + { + "Input": "a child want to buy 3 new toys . if there are 10 toys that satisfy the child ' s criteria , in how many different ways can the child select the 3 new toys if the order of selection does n ' t matter .", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 10.0\nn2 = 3.0\n\nanswer = scipy.special.comb(n1, n0)\nprint(answer)" + ], + "Output Answer": [ + "120" + ], + "split": "train" + }, + { + "Input": "out of 10 students in a class , 4 are from maharashtra , 3 are from karnataka , and 3 are from goa . four students are to be selected at random . what are the chances that at least one is from karnataka ?", + "Output Program": [ + "import math\nn0 = 10.0\nn1 = 4.0\nn2 = 3.0\nn3 = 3.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(4.0)))\nt2 = n0 - 4.0\nt3 = n0 - n2\nt4 = math.factorial(min(15, int(t2)))\nt5 = math.factorial(min(15, int(t3)))\nt6 = t3 - 4.0\nt7 = t0 / t4\nt8 = math.factorial(min(15, int(t6)))\nt9 = t7 / t1\nt10 = t5 / t8\nt11 = t10 / t1\nt12 = t9 - t11\nanswer = t12 / t9\nprint(answer)" + ], + "Output Answer": [ + "0.8333333333333334" + ], + "split": "train" + }, + { + "Input": "there are 15 boys and 10 girls in a class . if three students are selected at random , in how many ways that 1 girl and 2 boys are selected ?", + "Output Program": [ + "import scipy\nn0 = 15.0\nn1 = 10.0\nn2 = 1.0\nn3 = 2.0\nt0 = scipy.special.comb(n0, n3)\nt1 = scipy.special.comb(n1, n2)\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "1050" + ], + "split": "train" + }, + { + "Input": "in how many ways 6 boys and 4 girls can be seated in a row so that they are alternative ?", + "Output Program": [ + "import math\nn0 = 6.0\nn1 = 4.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "17280" + ], + "split": "train" + }, + { + "Input": "in how many ways can 22 books on english and 20 books on hindi be placed in a row on a shelf so that two books on hindi may not be together ?", + "Output Program": [ + "import math\nimport scipy\nn0 = 22.0\nn1 = 20.0\nt0 = n0 - 1.0\nt1 = scipy.special.comb(t0, 3.0)\nt2 = scipy.special.comb(t0, 1.0)\nt3 = math.factorial(int(max(t0, 2.0))) / math.factorial(int(max(t0, 2.0) - min(t0, 2.0))) # find all permutations\nt4 = t1 + t3\nanswer = t4 + t2\nprint(answer)" + ], + "Output Answer": [ + "1771" + ], + "split": "train" + }, + { + "Input": "a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability j that the first ball picked up is odd numbered ?", + "Output Program": [ + "n0 = 100.0\nn1 = 1.0\nn2 = 100.0\nn3 = 3.0\nn4 = 3.0\n\nanswer = 2.0 / n3\nprint(answer)" + ], + "Output Answer": [ + "0.6666666666666666" + ], + "split": "train" + }, + { + "Input": "a bag contains 3 white marbles and 3 black marbles . if each of 3 girls and 3 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 3.0\nn2 = 3.0\nn3 = 3.0\nt0 = n0 + n0\nt1 = scipy.special.comb(t0, n0)\nanswer = 2.0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.1" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 600 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 10.0\nn3 = 600.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "437.40000000000003" + ], + "split": "train" + }, + { + "Input": "five 6 faced dice are thrown together . the probability that all the three show the same number on them is ?", + "Output Program": [ + "n0 = 6.0\nt0 = 1.0 / n0\nt1 = t0 * t0\nt2 = t0 * t1\nanswer = t0 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.0007716049382716049" + ], + "split": "train" + }, + { + "Input": "there are 3 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed .", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 2.0\nt0 = scipy.special.comb(n0, n1)\nanswer = t0 + t0\nprint(answer)" + ], + "Output Answer": [ + "6" + ], + "split": "train" + }, + { + "Input": "if you roll one fair 6 - sided die , what is the probability that the number is even or less than 3 ?", + "Output Program": [ + "n0 = 6.0\nn1 = 3.0\nt0 = 3.0 / n0\nt1 = 1.0 / n0\nanswer = t0 + t1\nprint(answer)" + ], + "Output Answer": [ + "0.6666666666666666" + ], + "split": "train" + }, + { + "Input": "a box contains 20 electric bulbs , out of which 4 are defective . two bulbs are chosen at random form this box . the probability that at least one of these is defective is :", + "Output Program": [ + "import scipy\nn0 = 20.0\nn1 = 4.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, 2.0)\nt3 = t2 / t0\nanswer = 1.0 - t3\nprint(answer)" + ], + "Output Answer": [ + "0.368421052631579" + ], + "split": "train" + }, + { + "Input": "6 students are equally divided into 3 groups , then , the 3 groups were assigned to 3 different topics . how many different arrangements w are possible ?", + "Output Program": [ + "import math\nn0 = 6.0\nn1 = 3.0\nn2 = 3.0\nn3 = 3.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nt2 = n0 - 2.0\nt3 = math.factorial(min(15, int(t2)))\nt4 = t0 / t3\nt5 = t4 / 2.0\nanswer = t5 * t1\nprint(answer)" + ], + "Output Answer": [ + "90" + ], + "split": "train" + }, + { + "Input": "how many different ways can 2 students be seated in a row of 8 desks , so that there is always at least one empty desk between the students ?", + "Output Program": [ + "import math\nn0 = 2.0\nn1 = 8.0\nt0 = n1 - 1.0\nanswer = math.factorial(int(max(t0, n0))) / math.factorial(int(max(t0, n0) - min(t0, n0))) # find all permutations\nprint(answer)" + ], + "Output Answer": [ + "42" + ], + "split": "train" + }, + { + "Input": "a number is selected at random from the first 30 natural numbers . what is the probability that the number is a multiple of either 4 or 15 ?", + "Output Program": [ + "import math\nn0 = 30.0\nn1 = 4.0\nn2 = 15.0\nt0 = n0 / n2\nt1 = n0 / n1\nt2 = math.floor(t1)\nt3 = t0 + t2\nanswer = t3 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.3" + ], + "split": "train" + }, + { + "Input": "in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what is the probability that the extracted ball would turnon to be white , if all possible hypothesis concerning thecolor of theballs that initiallyin the bag were equally possible ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nt0 = n0 + n1\nt1 = 1.0 / 2.0\nt2 = n1 / t0\nt3 = n0 / t0\nt4 = t2 + t3\nt5 = t4 + t1\nt6 = n1 + t5\nanswer = t6 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.625" + ], + "split": "train" + }, + { + "Input": "wink , inc . follows a certain procedure that requires two tasks to be finished independently in order for a job to be done . on any given day , there is a 3 / 8 probability that task 1 will be completed on time , and a 3 / 5 probability that task 2 will be completed on time . on a certain day , what is the probability that task 1 will be completed on time , but task 2 will not ?", + "Output Program": [ + "n0 = 3.0\nn1 = 8.0\nn2 = 1.0\nn3 = 3.0\nn4 = 5.0\nn5 = 2.0\nn6 = 1.0\nn7 = 2.0\nt0 = n0 / n1\nt1 = n3 / n4\nt2 = n2 - t1\nanswer = t0 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.15000000000000002" + ], + "split": "train" + }, + { + "Input": "a box contains 24 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is", + "Output Program": [ + "import scipy\nn0 = 24.0\nn1 = 4.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, 2.0)\nt3 = t2 / t0\nanswer = 1.0 - t3\nprint(answer)" + ], + "Output Answer": [ + "0.3115942028985508" + ], + "split": "train" + }, + { + "Input": "a coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed . if 8 possible outcomes are equally likely . find the probability that the die shows a number greater than 4 , if it is known that the first throw of the coin results in a tail", + "Output Program": [ + "n0 = 8.0\nn1 = 4.0\nt0 = 2.0 / n0\nt1 = 6.0 / n0\nanswer = t0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "train" + }, + { + "Input": "what is the probability that jamie and tom will get selected for a job they both applied for in a company , given that their chances of getting selected is 2 / 3 and 5 / 7 respectively ?", + "Output Program": [ + "n0 = 2.0\nn1 = 3.0\nn2 = 5.0\nn3 = 7.0\nt0 = n0 / n1\nt1 = n2 / n3\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.47619047619047616" + ], + "split": "train" + }, + { + "Input": "a coin is tossed 6 times . what is the probability of getting exactly 5 heads ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 5.0\nt0 = scipy.special.comb(n0, n1)\nt1 = 1.0 / 2.0\nt2 = t0 * t1\nt3 = t1**min(n1, 5)\nanswer = t2 * t3\nprint(answer)" + ], + "Output Answer": [ + "0.09375" + ], + "split": "train" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 400 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 10.0\nn3 = 400.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "291.6" + ], + "split": "train" + }, + { + "Input": "a bag contains 6 black and 5 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ?", + "Output Program": [ + "import math\nn0 = 6.0\nn1 = 5.0\nt0 = n0 + n1\nt1 = math.factorial(min(15, int(n1)))\nt2 = math.factorial(min(15, int(2.0)))\nt3 = math.factorial(min(15, int(n0)))\nt4 = n1 - 2.0\nt5 = n0 - 2.0\nt6 = math.factorial(min(15, int(t4)))\nt7 = math.factorial(min(15, int(t5)))\nt8 = math.factorial(min(15, int(t0)))\nt9 = t0 - 2.0\nt10 = t1 / t6\nt11 = t3 / t7\nt12 = math.factorial(min(15, int(t9)))\nt13 = t10 / t2\nt14 = t11 / t2\nt15 = t8 / t12\nt16 = t13 + t14\nt17 = t15 / t2\nanswer = t16 / t17\nprint(answer)" + ], + "Output Answer": [ + "0.45454545454545453" + ], + "split": "train" + }, + { + "Input": "in how many ways 4 boys and 2 girls can be seated in a row so that they are alternate .", + "Output Program": [ + "import math\nn0 = 4.0\nn1 = 2.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "48" + ], + "split": "train" + }, + { + "Input": "a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability t that all 3 jellybeans removed from the bag are blue ?", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 10.0\nn2 = 3.0\nn3 = 3.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, n2)\nt2 = scipy.special.comb(t0, n2)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.10526315789473684" + ], + "split": "train" + }, + { + "Input": "a meeting has to be conducted with 4 managers . find the number of ways in which the managers be selected from among 7 managers , if 2 managers will not attend the meeting together ?", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 7.0\nn2 = 2.0\nt0 = scipy.special.comb(n1, n0)\nt1 = n1 - n2\nt2 = scipy.special.comb(t1, n2)\nanswer = t0 - t2\nprint(answer)" + ], + "Output Answer": [ + "25" + ], + "split": "train" + }, + { + "Input": "there are 6 books on a shelf , of which 2 are paperbacks and 4 are hardbacks . how many possible selections of 4 books from this shelf include at least one paperback ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 2.0\nn2 = 4.0\nn3 = 4.0\nt0 = scipy.special.comb(n0, n3)\nt1 = scipy.special.comb(n2, n3)\nanswer = t0 - t1\nprint(answer)" + ], + "Output Answer": [ + "14" + ], + "split": "train" + }, + { + "Input": "a jar contains 6 black and 2 white balls . if you pick two balls at the same time , what ' s the probability that one ball is black and one is white ?", + "Output Program": [ + "n0 = 6.0\nn1 = 2.0\nt0 = n0 + n1\nt1 = n0 / t0\nt2 = t0 - 1.0\nt3 = n1 / t2\nanswer = t1 * t3\nprint(answer)" + ], + "Output Answer": [ + "0.21428571428571427" + ], + "split": "train" + }, + { + "Input": "a father purchased dress for his 3 daughters . the dresses are of same color but diff size and they are kept in dark room . what is probability that all the 3 will not choose their own dress ?", + "Output Program": [ + "n0 = 65.0\nn1 = 2.0\nn2 = 29.0\nn3 = 5.0\nt0 = n2 * n3\nt1 = n0 - n2\nt2 = t0 / t1\nt3 = n0 * t2\nanswer = t3 / t2\nprint(answer)" + ], + "Output Answer": [ + "65" + ], + "split": "train" + }, + { + "Input": "how many different ways can 2 students be seated in a row of 3 desks , so that there is always at least one empty desk between the students ?", + "Output Program": [ + "import math\nn0 = 2.0\nn1 = 3.0\nt0 = n1 - 1.0\nanswer = math.factorial(int(max(t0, n0))) / math.factorial(int(max(t0, n0) - min(t0, n0))) # find all permutations\nprint(answer)" + ], + "Output Answer": [ + "2" + ], + "split": "train" + }, + { + "Input": "in a throw of dice what is the probability of ge \u00e6 \u00ab ng number greater than 2", + "Output Program": [ + "n0 = 2.0\nt0 = 6.0 - n0\nanswer = t0 / 6.0\nprint(answer)" + ], + "Output Answer": [ + "0.6666666666666666" + ], + "split": "train" + }, + { + "Input": "the probability that a number selected at random from first 50 natural numbers is a composite number is", + "Output Program": [ + "n0 = 50.0\nt0 = 3.0 * 6.0\nt1 = t0 - 1.0\nt2 = t1 * 2.0\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.68" + ], + "split": "train" + }, + { + "Input": "there are 5 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 4.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(t0, 2.0)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.2777777777777778" + ], + "split": "train" + }, + { + "Input": "joe has candys on his pocket , 4 of those candies have lemon flavor , and 3 have caramel flavor . if joe take one , what is the probability that he picks one caramel flavor ?", + "Output Program": [ + "n0 = 4.0\nn1 = 3.0\nt0 = n0 + n1\nanswer = n1 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.42857142857142855" + ], + "split": "train" + }, + { + "Input": "if a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exactly once in 3 consecutive flips ?", + "Output Program": [ + "n0 = 3.0\nt0 = 1.0 / 2.0\nt1 = t0**min(n0, 5)\nanswer = n0 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.375" + ], + "split": "train" + }, + { + "Input": "a coin is tossed 3 times . what is the probability of getting exactly 2 heads ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 2.0\nt0 = scipy.special.comb(n0, n1)\nt1 = 1.0 / 2.0\nt2 = t0 * t1\nt3 = t1**min(n1, 5)\nanswer = t2 * t3\nprint(answer)" + ], + "Output Answer": [ + "0.375" + ], + "split": "train" + }, + { + "Input": "there are 8 books on a shelf , of which 2 are paperbacks and 6 are hardbacks . how many possible selections of 3 books from this shelf include at least one paperback ?", + "Output Program": [ + "import scipy\nn0 = 8.0\nn1 = 2.0\nn2 = 6.0\nn3 = 3.0\nt0 = scipy.special.comb(n0, n3)\nt1 = scipy.special.comb(n2, n3)\nanswer = t0 - t1\nprint(answer)" + ], + "Output Answer": [ + "36" + ], + "split": "train" + }, + { + "Input": "a jar contains only red , yellow , and orange marbles . if there are 3 red , 5 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 5.0\nn2 = 4.0\nn3 = 3.0\nn4 = 2.0\nn5 = 1.0\nt0 = n1 + n2\nt1 = scipy.special.comb(n1, n4)\nt2 = scipy.special.comb(n0, n5)\nt3 = n0 + t0\nt4 = t1 * t2\nt5 = scipy.special.comb(t3, n0)\nanswer = t4 / t5\nprint(answer)" + ], + "Output Answer": [ + "0.13636363636363635" + ], + "split": "train" + }, + { + "Input": "a basket contains 7 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ?", + "Output Program": [ + "import scipy\nn0 = 7.0\nn1 = 1.0\nn2 = 2.0\nn3 = 2.0\nt0 = scipy.special.comb(n0, n2)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, n1)\nanswer = t2 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.2857142857142857" + ], + "split": "train" + }, + { + "Input": "the only items of clothing at michelle \u2019 s boutique are 4 shirts , 8 pairs of pants and 18 ties . what is the probability of randomly selecting 2 ties out of all the items in michelle \u2019 s boutique ?", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 8.0\nn2 = 18.0\nn3 = 2.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n2, 2.0)\nt2 = n2 + t0\nt3 = scipy.special.comb(t2, 2.0)\nanswer = t1 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.35172413793103446" + ], + "split": "train" + }, + { + "Input": "what is the probability of randomly selecting an arrangement of the letters of the word ` ` mediterranean ' ' in which the first letter is e and the last letter is r ?", + "Output Program": [ + "t0 = 26.0 / 2.0\nt1 = 2.0 * 3.0\nt2 = t1 * 2.0\nt3 = t0 * t2\nanswer = t1 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.038461538461538464" + ], + "split": "train" + }, + { + "Input": "there are 9 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed .", + "Output Program": [ + "import scipy\nn0 = 9.0\nn1 = 2.0\nt0 = scipy.special.comb(n0, n1)\nanswer = t0 + t0\nprint(answer)" + ], + "Output Answer": [ + "72" + ], + "split": "train" + }, + { + "Input": "in a full deck of 52 cards , there are 13 spades . a game is played whereby a card from is chosen at random from the deck . if the card is a spade , a winner is declared . if the card is not a spade , it is returned to the deck , the deck is reshuffled and another card is drawn . this process is repeated until a spade is drawn . what is the probability that one will pick the first spade on the second draw or later ?", + "Output Program": [ + "n0 = 52.0\nn1 = 13.0\nt0 = n1 / n0\nanswer = 1.0 - t0\nprint(answer)" + ], + "Output Answer": [ + "0.75" + ], + "split": "train" + }, + { + "Input": "a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability w that all 3 jellybeans removed from the bag are blue ?", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 10.0\nn2 = 3.0\nn3 = 3.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, n2)\nt2 = scipy.special.comb(t0, n2)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.10526315789473684" + ], + "split": "train" + }, + { + "Input": "in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what isthe probability that theextracted ball would turn on to be white , if all possible hypothesis concerning the color of the balls that initially in the bag were equally possible ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nt0 = n0 + n1\nt1 = n1 / 2.0\nt2 = n1 / t0\nt3 = n0 / t0\nt4 = t2 + t3\nt5 = t4 + t1\nt6 = n1 + t5\nanswer = t6 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.625" + ], + "split": "train" + }, + { + "Input": "if a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land tails up exactly once in 3 consecutive flips ?", + "Output Program": [ + "n0 = 3.0\nt0 = 1.0 / 2.0\nt1 = t0**min(3.0, 5)\nanswer = n0 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.375" + ], + "split": "train" + }, + { + "Input": "jane and thomas are among the 6 people from which a committee of 3 people is to be selected . how many different possible committees of 3 people can be selected from these 6 people if at least one of either jane or thomas is to be selected ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 3.0\nn2 = 3.0\nn3 = 6.0\nt0 = scipy.special.comb(n0, 1.0)\nt1 = scipy.special.comb(4.0, 1.0)\nt2 = t0 + t0\nanswer = t2 + t1\nprint(answer)" + ], + "Output Answer": [ + "16" + ], + "split": "train" + }, + { + "Input": "there is a 30 % chance jen will visit chile this year , while there is a 50 % chance that she will visit madagascar this year . what is the probability that jen will visit either chile or madagascar this year , but not both ?", + "Output Program": [ + "n0 = 30.0\nn1 = 50.0\nt0 = 2.0 * 5.0\nt1 = t0**min(2.0, 5)\nt2 = n1 / t1\nt3 = n0 / t1\nt4 = t1 - n0\nt5 = t4 / t1\nt6 = t3 * t2\nt7 = t5 * t2\nt8 = t7 + t6\nanswer = t8 * t1\nprint(answer)" + ], + "Output Answer": [ + "50" + ], + "split": "train" + }, + { + "Input": "30 ladies and 60 gentlemen are present at a party . there are 23 couples among them . if a lady and a gentleman is selected at random , what is the probability that they will be a couple ?", + "Output Program": [ + "n0 = 30.0\nn1 = 60.0\nn2 = 23.0\nt0 = n0 * n1\nanswer = n2 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.012777777777777779" + ], + "split": "train" + }, + { + "Input": "at a conference table , 6 employees gather around a table . one of the employees is the manager and he sits at the head of the table . two of the employees sit on either side of the table . how many different seating arrangements can be made with these 6 employees ?", + "Output Program": [ + "import math\nn0 = 6.0\nn1 = 6.0\nt0 = n0 - 1.0\nanswer = math.factorial(min(15, int(t0)))\nprint(answer)" + ], + "Output Answer": [ + "120" + ], + "split": "train" + }, + { + "Input": "a bag contains 6 white marbles and 6 black marbles . if each of 6 girls and 6 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 6.0\nn2 = 6.0\nn3 = 6.0\nt0 = n0 + n0\nt1 = scipy.special.comb(t0, n0)\nanswer = 2.0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.0021645021645021645" + ], + "split": "train" + }, + { + "Input": "a bag contains 4 white marbles and 4 black marbles . if each of 4 girls and 4 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ?", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 4.0\nn2 = 4.0\nn3 = 4.0\nt0 = n0 + n0\nt1 = scipy.special.comb(t0, n0)\nanswer = 2.0 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.02857142857142857" + ], + "split": "train" + }, + { + "Input": "a basket contains 8 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ?", + "Output Program": [ + "import scipy\nn0 = 8.0\nn1 = 1.0\nn2 = 2.0\nn3 = 2.0\nt0 = scipy.special.comb(n0, n2)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, n1)\nanswer = t2 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.25" + ], + "split": "train" + }, + { + "Input": "a life insurance company insured 25,000 young boys , 14,000 young girls and 16,000 young adults . the probability of death within 10 years of a young boy , young girl and a young adult are 0.02 , 0.03 and 0.15 respectively . one insured person dies . what is the probability that the dead person is a young boy ?", + "Output Program": [ + "n0 = 25000.0\nn1 = 14000.0\nn2 = 16000.0\nn3 = 10.0\nn4 = 0.02\nn5 = 0.03\nn6 = 0.15\nt0 = 3.0 + 4.0\nt1 = 5.0**min(2.0, 5)\nt2 = 4.0**min(2.0, 5)\nt3 = n4 * t1\nt4 = t0 * 2.0\nt5 = n6 * t2\nt6 = n5 * t4\nt7 = t3 + t6\nt8 = t7 + t5\nanswer = t3 / t8\nprint(answer)" + ], + "Output Answer": [ + "0.15060240963855423" + ], + "split": "train" + }, + { + "Input": "a , b , c , d , e , f , g , h sitting in a row what is the probability that a , b , d are sitting together ?", + "Output Program": [ + "import math\nt0 = 4.0 + 4.0\nt1 = math.factorial(min(15, int(3.0)))\nt2 = math.factorial(min(15, int(t0)))\nt3 = t0 - 3.0\nt4 = t3 + 1.0\nt5 = math.factorial(min(15, int(t4)))\nt6 = t5 * t1\nanswer = t6 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.10714285714285714" + ], + "split": "train" + }, + { + "Input": "the membership of a committee consists of 3 english teachers , 4 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee \u2019 s report , what is the probability that the two members selected will both be maths teachers ?", + "Output Program": [ + "n0 = 3.0\nn1 = 4.0\nn2 = 2.0\nn3 = 2.0\nt0 = n0 + n1\nt1 = n2 + t0\nt2 = n0 / t1\nt3 = t1 - 1.0\nt4 = n2 / t3\nanswer = t2 * t4\nprint(answer)" + ], + "Output Answer": [ + "0.08333333333333333" + ], + "split": "train" + }, + { + "Input": "a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability c that out of the two , one is white and one is blue ?", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n0 + n1\nt1 = n1 + 2.0\nt2 = n0 * n1\nt3 = t0 * t1\nt4 = t3 / 2.0\nanswer = t2 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.5357142857142857" + ], + "split": "train" + }, + { + "Input": "two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability v that jane will be on the same committee as albert ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 3.0\nt0 = scipy.special.comb(4.0, 1.0)\nt1 = scipy.special.comb(n0, n1)\nt2 = 5.0 * 5.0\nt3 = t0 * 2.0\nt4 = t2 * 4.0\nt5 = t3 / t1\nanswer = t5 * t4\nprint(answer)" + ], + "Output Answer": [ + "40" + ], + "split": "train" + }, + { + "Input": "wink , inc . follows a certain procedure that requires two tasks to be finished independently in order for a job to be done . on any given day , there is a 2 / 3 probability that task 1 will be completed on time , and a 3 / 5 probability that task 2 will be completed on time . on a certain day , what is the probability that task 1 will be completed on time , but task 2 will not ?", + "Output Program": [ + "n0 = 2.0\nn1 = 3.0\nn2 = 1.0\nn3 = 3.0\nn4 = 5.0\nn5 = 2.0\nn6 = 1.0\nn7 = 2.0\nt0 = n0 / n1\nt1 = n3 / n4\nt2 = n2 - t1\nanswer = t0 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.26666666666666666" + ], + "split": "train" + }, + { + "Input": "if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first 4 flips but not on the last flip ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 5.0\nn3 = 4.0\nt0 = n0 / n1\nanswer = t0**min(n2, 5)\nprint(answer)" + ], + "Output Answer": [ + "0.03125" + ], + "split": "train" + }, + { + "Input": "the membership of a committee consists of 3 english teachers , 4 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee \u2019 s report , what is the probability that the two members selected will both be english teachers ?", + "Output Program": [ + "n0 = 3.0\nn1 = 4.0\nn2 = 2.0\nn3 = 2.0\nt0 = n0 + n1\nt1 = n2 + t0\nt2 = n0 / t1\nt3 = t1 - 1.0\nt4 = n2 / t3\nanswer = t2 * t4\nprint(answer)" + ], + "Output Answer": [ + "0.08333333333333333" + ], + "split": "dev" + }, + { + "Input": "when two dice are tossed , what is the probability that the total score is a prime number ?", + "Output Program": [ + "t0 = 2.0 * 6.0\nt1 = 6.0 * 6.0\nt2 = t0 + 3.0\nanswer = t2 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.4166666666666667" + ], + "split": "dev" + }, + { + "Input": "how many different ways can 2 students be seated in a row of 5 desks , so that there is always at least one empty desk between the students ?", + "Output Program": [ + "import math\nn0 = 2.0\nn1 = 5.0\nt0 = n1 - 1.0\nanswer = math.factorial(int(max(t0, n0))) / math.factorial(int(max(t0, n0) - min(t0, n0))) # find all permutations\nprint(answer)" + ], + "Output Answer": [ + "12" + ], + "split": "dev" + }, + { + "Input": "a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability e that all 3 jellybeans removed from the bag are blue ?", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 10.0\nn2 = 3.0\nn3 = 3.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, n2)\nt2 = scipy.special.comb(t0, n2)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.10526315789473684" + ], + "split": "dev" + }, + { + "Input": "a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability r that the first ball picked up is odd numbered ?", + "Output Program": [ + "n0 = 100.0\nn1 = 1.0\nn2 = 100.0\nn3 = 3.0\nn4 = 3.0\n\nanswer = 2.0 / n3\nprint(answer)" + ], + "Output Answer": [ + "0.6666666666666666" + ], + "split": "dev" + }, + { + "Input": "there are 15 balls in a jar . you take out 3 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 / 3 . how many blue balls were there in the beginning ?", + "Output Program": [ + "n0 = 15.0\nn1 = 3.0\nn2 = 1.0\nn3 = 3.0\nt0 = n0 - n1\nt1 = t0 / n3\nanswer = n1 + t1\nprint(answer)" + ], + "Output Answer": [ + "7" + ], + "split": "dev" + }, + { + "Input": "there are 7 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed .", + "Output Program": [ + "import scipy\nn0 = 7.0\nn1 = 2.0\nt0 = scipy.special.comb(n0, n1)\nanswer = t0 + t0\nprint(answer)" + ], + "Output Answer": [ + "42" + ], + "split": "dev" + }, + { + "Input": "a bag consists of 20 marbles , of which 5 are blue , 7 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ?", + "Output Program": [ + "n0 = 20.0\nn1 = 5.0\nn2 = 7.0\nt0 = n1 + n2\nt1 = n0 - t0\nt2 = n2 + t1\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.75" + ], + "split": "dev" + }, + { + "Input": "what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular decagon ( 10 - sided polygon ) ?", + "Output Program": [ + "n0 = 10.0\nt0 = n0 - 3.0\nt1 = n0 * t0\nt2 = t1 / 2.0\nanswer = n0 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.2857142857142857" + ], + "split": "dev" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 4 . for a group of 200 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 4.0\nn3 = 200.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "84.375" + ], + "split": "dev" + }, + { + "Input": "in a throw of dice what is the probability of getting number greater than 5", + "Output Program": [ + "n0 = 5.0\nt0 = 6.0 - n0\nanswer = t0 / 6.0\nprint(answer)" + ], + "Output Answer": [ + "0.16666666666666666" + ], + "split": "dev" + }, + { + "Input": "when two dice are rolled , what is the probability that the sum of the two numbers is 6 ?", + "Output Program": [ + "n0 = 6.0\nt0 = n0 * n0\nt1 = n0 - 1.0\nanswer = t1 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.1388888888888889" + ], + "split": "dev" + }, + { + "Input": "a box contains 22 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is", + "Output Program": [ + "import scipy\nn0 = 22.0\nn1 = 4.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, 2.0)\nt3 = t2 / t0\nanswer = 1.0 - t3\nprint(answer)" + ], + "Output Answer": [ + "0.33766233766233766" + ], + "split": "dev" + }, + { + "Input": "a box contains 21 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is", + "Output Program": [ + "import scipy\nn0 = 21.0\nn1 = 4.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, 2.0)\nt3 = t2 / t0\nanswer = 1.0 - t3\nprint(answer)" + ], + "Output Answer": [ + "0.35238095238095235" + ], + "split": "dev" + }, + { + "Input": "kim has 3 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 6 shoes , what is the probability that she will select 2 shoes of the same color ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 2.0\nn2 = 6.0\nn3 = 2.0\nt0 = scipy.special.comb(n2, n1)\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.2" + ], + "split": "dev" + }, + { + "Input": "a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many ways can the test be completed if every question is unanswered ?", + "Output Program": [ + "n0 = 4.0\nn1 = 5.0\n\nanswer = n1**min(n0, 5)\nprint(answer)" + ], + "Output Answer": [ + "625" + ], + "split": "dev" + }, + { + "Input": "an urn contains 5 red , 6 blue and 8 green balls . 4 balls are randomly selected from the urn , find the probability that the drawn ball are 1 blue and 3 red ?", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 6.0\nn2 = 8.0\nn3 = 4.0\nn4 = 1.0\nn5 = 3.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n1, n4)\nt2 = scipy.special.comb(n0, n5)\nt3 = n2 + t0\nt4 = t1 * t2\nt5 = scipy.special.comb(t3, n3)\nanswer = t4 / t5\nprint(answer)" + ], + "Output Answer": [ + "0.015479876160990712" + ], + "split": "dev" + }, + { + "Input": "in a department store prize box , 40 % of the notes give the winner a dreamy vacation ; the other notes are blank . what is the approximate probability that 3 out of 5 people that draw the notes one after the other , and immediately return their note into the box get a dreamy vacation ?", + "Output Program": [ + "import scipy\nn0 = 40.0\nn1 = 3.0\nn2 = 5.0\nt0 = scipy.special.comb(n2, n1)\nt1 = 3.0 / n2\nt2 = 2.0 / n2\nt3 = t1**min(2.0, 5)\nt4 = t2**min(3.0, 5)\nt5 = t0 * t3\nanswer = t5 * t4\nprint(answer)" + ], + "Output Answer": [ + "0.23040000000000002" + ], + "split": "dev" + }, + { + "Input": "a , b , c , d , e , f , g , h sitting in a row what is the probability that a , b , c sitting together ?", + "Output Program": [ + "import math\nt0 = 4.0 + 4.0\nt1 = math.factorial(min(15, int(3.0)))\nt2 = math.factorial(min(15, int(t0)))\nt3 = t0 - 3.0\nt4 = t3 + 1.0\nt5 = math.factorial(min(15, int(t4)))\nt6 = t5 * t1\nanswer = t6 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.10714285714285714" + ], + "split": "dev" + }, + { + "Input": "from a group of 3 boys and 3 girls , 4 children are to be randomly selected . what is the probability that equal numbers of boys and girls will be selected ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 3.0\nn2 = 4.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(3.0, 2.0)\nt3 = scipy.special.comb(t0, 4.0)\nt4 = t1 * t2\nanswer = t4 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.6" + ], + "split": "dev" + }, + { + "Input": "4 dices are thrown at the same time . what is the probability of getting only 2 dices showing the same face ?", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 2.0\nt0 = scipy.special.comb(n0, n1)\nt1 = 6.0**min(4.0, 5)\nt2 = t0 * 6.0\nt3 = t2 * 5.0\nt4 = t3 * 4.0\nanswer = t4 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.5555555555555556" + ], + "split": "dev" + }, + { + "Input": "the odds in favour of an events are 3 : 5 . the probability of occurrence of the event is", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n0 + n1\nanswer = n0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.375" + ], + "split": "dev" + }, + { + "Input": "3 houses are available in a new flat . 3 persons apply for the houses . each applies for 1 house without consulting others . the probability that all the 3 apply for the same house is ?", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 3.0\nn2 = 1.0\nn3 = 3.0\nt0 = scipy.special.comb(n0, n2)\nt1 = t0 * t0\nanswer = n2 / t1\nprint(answer)" + ], + "Output Answer": [ + "0.1111111111111111" + ], + "split": "dev" + }, + { + "Input": "the chance of rain on any given day in tel - aviv is 50 % . what is the probability w that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ?", + "Output Program": [ + "import scipy\nn0 = 50.0\nn1 = 4.0\nn2 = 6.0\nt0 = scipy.special.comb(n2, n1)\nt1 = 1.0 / 2.0\nt2 = t1**min(4.0, 5)\nt3 = t1**min(2.0, 5)\nt4 = t2 * t3\nt5 = 1 / t4\nanswer = t0 / t5\nprint(answer)" + ], + "Output Answer": [ + "0.234375" + ], + "split": "dev" + }, + { + "Input": "the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability t that the next couple the maitre ' d seats will not order dessert ?", + "Output Program": [ + "n0 = 60.0\nn1 = 20.0\nt0 = 2.0 * 5.0\nt1 = t0 * t0\nt2 = n0 / t1\nt3 = n1 / t1\nt4 = 1.0 - t3\nt5 = t2 / t4\nt6 = 1.0 - t5\nanswer = t1 * t6\nprint(answer)" + ], + "Output Answer": [ + "25.00000000000001" + ], + "split": "dev" + }, + { + "Input": "the probability that a number selected at random from the first 50 natural numbers is a composite number is", + "Output Program": [ + "n0 = 50.0\nt0 = 3.0 * 6.0\nt1 = t0 - 1.0\nt2 = t1 * 2.0\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.68" + ], + "split": "dev" + }, + { + "Input": "two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability that jane will be on the same committee as albert ?", + "Output Program": [ + "import scipy\nn0 = 6.0\nn1 = 3.0\nt0 = scipy.special.comb(4.0, 1.0)\nt1 = scipy.special.comb(n0, n1)\nt2 = 5.0 * 5.0\nt3 = t0 * 2.0\nt4 = t2 * 4.0\nt5 = t3 / t1\nanswer = t5 * t4\nprint(answer)" + ], + "Output Answer": [ + "40" + ], + "split": "dev" + }, + { + "Input": "a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability w that out of the two , one is white and one is blue ?", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n0 + n1\nt1 = n1 + 2.0\nt2 = n0 * n1\nt3 = t0 * t1\nt4 = t3 / 2.0\nanswer = t2 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.5357142857142857" + ], + "split": "dev" + }, + { + "Input": "a man & his wife appear in an interview for 2 vacancies in the same post . the probability of husband ' s selection is 1 / 7 & the probability of wife ' s selection is 1 / 5 . what is the probability that only 1 of them is selected ?", + "Output Program": [ + "n0 = 2.0\nn1 = 1.0\nn2 = 7.0\nn3 = 1.0\nn4 = 5.0\nn5 = 1.0\nt0 = n1 / n4\nt1 = n1 / n2\nt2 = n1 - t1\nt3 = n1 - t0\nt4 = t0 * t2\nt5 = t1 * t3\nanswer = t4 + t5\nprint(answer)" + ], + "Output Answer": [ + "0.28571428571428575" + ], + "split": "dev" + }, + { + "Input": "there are 25 balls in a jar . you take out 5 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 / 5 . how many blue balls were there in the beginning ?", + "Output Program": [ + "n0 = 25.0\nn1 = 5.0\nn2 = 1.0\nn3 = 5.0\nt0 = n0 - n1\nt1 = t0 / n3\nanswer = n1 + t1\nprint(answer)" + ], + "Output Answer": [ + "9" + ], + "split": "dev" + }, + { + "Input": "tom and marry each have a bag of marbles , each of which contains 4 red marbles , 10 blue marbles , and 6 yellow marbles . if tom and marry each select one marble from their respective bags , what is the probability that either tom or marry select a blue marble ?", + "Output Program": [ + "n0 = 4.0\nn1 = 10.0\nn2 = 6.0\nt0 = n0 + n2\nt1 = n1 + t0\nt2 = n1 / t1\nt3 = t2 * t2\nanswer = 1.0 - t3\nprint(answer)" + ], + "Output Answer": [ + "0.75" + ], + "split": "dev" + }, + { + "Input": "a bag consists of 30 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ?", + "Output Program": [ + "n0 = 30.0\nn1 = 5.0\nn2 = 9.0\nt0 = n1 + n2\nt1 = n0 - t0\nt2 = n2 + t1\nanswer = t2 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.8333333333333334" + ], + "split": "dev" + }, + { + "Input": "in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 150 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?", + "Output Program": [ + "n0 = 3.0\nn1 = 1.0\nn2 = 10.0\nn3 = 150.0\nn4 = 3.0\nt0 = n1 / n2\nt1 = n1 - t0\nt2 = n3 * t1\nt3 = t2 * t1\nanswer = t3 * t1\nprint(answer)" + ], + "Output Answer": [ + "109.35000000000001" + ], + "split": "dev" + }, + { + "Input": "two letters from the word barkhint are selected at random . what is the probability that at least one of the selected letters is a consonant ?", + "Output Program": [ + "import scipy\nt0 = scipy.special.comb(2.0, 2.0)\nt1 = 2.0 * 4.0\nt2 = scipy.special.comb(t1, 2.0)\nt3 = t2 - t0\nanswer = t3 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.9642857142857143" + ], + "split": "dev" + }, + { + "Input": "the chance of rain on any given day in tel - aviv is 50 % . what is the probability z that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ?", + "Output Program": [ + "import scipy\nn0 = 50.0\nn1 = 4.0\nn2 = 6.0\nt0 = scipy.special.comb(n2, n1)\nt1 = 1.0 / 2.0\nt2 = t1**min(n1, 5)\nt3 = t1**min(2.0, 5)\nt4 = t2 * t3\nt5 = 1 / t4\nanswer = t0 / t5\nprint(answer)" + ], + "Output Answer": [ + "0.234375" + ], + "split": "dev" + }, + { + "Input": "a fair coin is tossed 5 times . what is the probability that it lands heads up at least twice ?", + "Output Program": [ + "import scipy\nn0 = 5.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = scipy.special.comb(n0, 3.0)\nt2 = scipy.special.comb(n0, 4.0)\nt3 = scipy.special.comb(n0, n0)\nt4 = 2.0**min(n0, 5)\nt5 = t0 + t1\nt6 = t5 + t2\nt7 = t6 + t3\nanswer = t7 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.8125" + ], + "split": "dev" + }, + { + "Input": "what is the probability that the sum of two dice will yield a 5 , and then when both are thrown again , their sum will again yield a 5 ? assume that each die has 4 sides with faces numbered 1 to 4 .", + "Output Program": [ + "n0 = 5.0\nn1 = 5.0\nn2 = 4.0\nn3 = 1.0\nn4 = 4.0\nt0 = n2**min(2.0, 5)\nt1 = n2 / t0\nanswer = t1 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.0625" + ], + "split": "dev" + }, + { + "Input": "in a throw of dice what is the probability of ge \u00e6 \u00ab ng number greater than 4", + "Output Program": [ + "n0 = 4.0\nt0 = 6.0 - n0\nanswer = t0 / 6.0\nprint(answer)" + ], + "Output Answer": [ + "0.3333333333333333" + ], + "split": "test" + }, + { + "Input": "what is the probability that the sum of two dice will yield a 6 , and then when both are thrown again , their sum will again yield a 6 ? assume that each die has 5 sides with faces numbered 1 to 5 .", + "Output Program": [ + "n0 = 6.0\nn1 = 6.0\nn2 = 5.0\nn3 = 1.0\nn4 = 5.0\nt0 = n2**min(2.0, 5)\nt1 = n2 / t0\nanswer = t1 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.04000000000000001" + ], + "split": "test" + }, + { + "Input": "in how many ways 4 boys and 4 girls can be seated in a row so that they are alternate .", + "Output Program": [ + "import math\nn0 = 4.0\nn1 = 4.0\nt0 = math.factorial(min(15, int(n0)))\nt1 = math.factorial(min(15, int(n1)))\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "576" + ], + "split": "test" + }, + { + "Input": "a basket contains 10 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ?", + "Output Program": [ + "import scipy\nn0 = 10.0\nn1 = 1.0\nn2 = 2.0\nn3 = 2.0\nt0 = scipy.special.comb(n0, n2)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, n1)\nanswer = t2 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.2" + ], + "split": "test" + }, + { + "Input": "a basket contains 9 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ?", + "Output Program": [ + "import scipy\nn0 = 9.0\nn1 = 1.0\nn2 = 2.0\nn3 = 2.0\nt0 = scipy.special.comb(n0, n2)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, n1)\nanswer = t2 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.2222222222222222" + ], + "split": "test" + }, + { + "Input": "in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 1 / 5 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and straight", + "Output Program": [ + "n0 = 1.0\nn1 = 5.0\nn2 = 1.0\nn3 = 2.0\nt0 = n2 / n3\nt1 = n0 / n1\nt2 = n2 - t1\nanswer = t0 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.4" + ], + "split": "test" + }, + { + "Input": "rhonda picked 2 pen from the table , if there were 7 pens on the table and 5 belongs to jill , what is the probability that the 2 pen she picked does not belong to jill ? .", + "Output Program": [ + "n0 = 2.0\nn1 = 7.0\nn2 = 5.0\nn3 = 2.0\nt0 = n1 - n2\nt1 = n1 - 1.0\nt2 = t0 / n1\nt3 = t0 - 1.0\nt4 = t3 / t1\nanswer = t2 * t4\nprint(answer)" + ], + "Output Answer": [ + "0.047619047619047616" + ], + "split": "test" + }, + { + "Input": "a box contains 25 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is", + "Output Program": [ + "import scipy\nn0 = 25.0\nn1 = 4.0\nt0 = scipy.special.comb(n0, 2.0)\nt1 = n0 - n1\nt2 = scipy.special.comb(t1, 2.0)\nt3 = t2 / t0\nanswer = 1.0 - t3\nprint(answer)" + ], + "Output Answer": [ + "0.30000000000000004" + ], + "split": "test" + }, + { + "Input": "how many different ways can 2 students be seated in a row of 4 desks , so that there is always at least one empty desk between the students ?", + "Output Program": [ + "import math\nn0 = 2.0\nn1 = 4.0\nt0 = n1 - 1.0\nanswer = math.factorial(int(max(t0, n0))) / math.factorial(int(max(t0, n0) - min(t0, n0))) # find all permutations\nprint(answer)" + ], + "Output Answer": [ + "6" + ], + "split": "test" + }, + { + "Input": "if two dice are thrown together , the probability of getting a doublet on the dice is", + "Output Program": [ + "t0 = 6.0 * 6.0\nanswer = 6.0 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.16666666666666666" + ], + "split": "test" + }, + { + "Input": "there are 13 clubs in a full deck of 52 cards . in a certain game , you pick a card from a standard deck of 52 cards . if the card is a club , you win . if the card is not a club , the person replaces the card to the deck , reshuffles , and draws again . the person keeps repeating that process until he picks a club , and the point is to measure how many draws it took before the person picked a club and , thereby , won . what is the probability that one will pick the first club on the forth draw or later ?", + "Output Program": [ + "n0 = 13.0\nn1 = 52.0\nn2 = 52.0\nt0 = n1 - n0\nt1 = t0 / n1\nt2 = t1 * t1\nanswer = t1 * t2\nprint(answer)" + ], + "Output Answer": [ + "0.421875" + ], + "split": "test" + }, + { + "Input": "if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 5.0\nn3 = 3.0\nn4 = 2.0\nt0 = n0 / n1\nanswer = t0**min(n2, 5)\nprint(answer)" + ], + "Output Answer": [ + "0.03125" + ], + "split": "test" + }, + { + "Input": "the membership of a committee consists of 3 english teachers , 4 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee \u2019 s report , what is the probability that the two members selected will both be social teachers ?", + "Output Program": [ + "n0 = 3.0\nn1 = 4.0\nn2 = 2.0\nn3 = 2.0\nt0 = n0 + n1\nt1 = n2 + t0\nt2 = n0 / t1\nt3 = t1 - 1.0\nt4 = n2 / t3\nanswer = t2 * t4\nprint(answer)" + ], + "Output Answer": [ + "0.08333333333333333" + ], + "split": "test" + }, + { + "Input": "there are 3 red shoes & 7 green shoes . if two of red shoes are drawn what is the probability of getting red shoes", + "Output Program": [ + "import scipy\nn0 = 3.0\nn1 = 7.0\nt0 = n0 + n1\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(t0, 2.0)\nanswer = t1 / t2\nprint(answer)" + ], + "Output Answer": [ + "0.06666666666666667" + ], + "split": "test" + }, + { + "Input": "what is the probability that a two digit number selected at random is a multiple of 3 and not a multiple of 5 ?", + "Output Program": [ + "n0 = 3.0\nn1 = 5.0\nt0 = n1 * 2.0\nt1 = n0 * 2.0\nt2 = n0 * n0\nt3 = n0 * t0\nt4 = t0 * t2\nt5 = t3 - t1\nanswer = t5 / t4\nprint(answer)" + ], + "Output Answer": [ + "0.26666666666666666" + ], + "split": "test" + }, + { + "Input": "there are 8 books on a shelf , of which 3 are paperbacks and 5 are hardbacks . how many possible selections of 4 books from this shelf include at least one paperback ?", + "Output Program": [ + "import scipy\nn0 = 8.0\nn1 = 3.0\nn2 = 5.0\nn3 = 4.0\nt0 = scipy.special.comb(n0, n3)\nt1 = scipy.special.comb(n2, n3)\nanswer = t0 - t1\nprint(answer)" + ], + "Output Answer": [ + "65" + ], + "split": "test" + }, + { + "Input": "denise is trying to open a safe whose combination she does not know . if the safe has 4000 possible combinations , and she can try 75 different possibilities , what is the probability that she does not pick the one correct combination .", + "Output Program": [ + "n0 = 4000.0\nn1 = 75.0\nt0 = n0 - n1\nanswer = t0 / n0\nprint(answer)" + ], + "Output Answer": [ + "0.98125" + ], + "split": "test" + }, + { + "Input": "there are 5 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed .", + "Output Program": [ + "import scipy\nn0 = 5.0\nn1 = 2.0\nt0 = scipy.special.comb(n0, n1)\nanswer = t0 + t0\nprint(answer)" + ], + "Output Answer": [ + "20" + ], + "split": "test" + }, + { + "Input": "x , y , and z are all unique numbers . if x is chosen randomly from the set { 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ?", + "Output Program": [ + "n0 = 10.0\nn1 = 11.0\nn2 = 20.0\nn3 = 21.0\nn4 = 22.0\nn5 = 23.0\nt0 = 1.0 / 2.0\nt1 = 1.0 / 4.0\nanswer = t0 * t1\nprint(answer)" + ], + "Output Answer": [ + "0.125" + ], + "split": "test" + }, + { + "Input": "from a pack of cards , two cards are drawn one after the other , with replacement . what is the probability that the first card is a club and the second card is a red king ?", + "Output Program": [ + "t0 = 2.0 / 52.0\nt1 = 3.0 * 4.0\nt2 = t1 + 1.0\nt3 = t2 / 52.0\nanswer = t3 * t0\nprint(answer)" + ], + "Output Answer": [ + "0.009615384615384616" + ], + "split": "test" + }, + { + "Input": "a man speaks truth 3 out of 4 times . he throws a die and reports it to be a 6 . what is the probability of it being a 6 ?", + "Output Program": [ + "n0 = 3.0\nn1 = 4.0\nn2 = 6.0\nn3 = 6.0\nt0 = n0 / n1\nt1 = 1.0 / n2\nt2 = 1.0 / 4.0\nt3 = 5.0 / n2\nt4 = t0 * t1\nt5 = t2 * t3\nt6 = t4 + t5\nanswer = t4 / t6\nprint(answer)" + ], + "Output Answer": [ + "0.37499999999999994" + ], + "split": "test" + }, + { + "Input": "if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is - .", + "Output Program": [ + "t0 = 3.0 / 52.0\nt1 = 52.0 / 4.0\nt2 = t1 / 52.0\nanswer = t0 + t2\nprint(answer)" + ], + "Output Answer": [ + "0.3076923076923077" + ], + "split": "test" + }, + { + "Input": "from a group of 4 boys and 4 girls , 4 children are to be randomly selected . what is the probability that 2 boys and 2 girls will be selected ?", + "Output Program": [ + "import scipy\nn0 = 4.0\nn1 = 4.0\nn2 = 4.0\nn3 = 2.0\nn4 = 2.0\nt0 = n0 + n0\nt1 = scipy.special.comb(n0, 2.0)\nt2 = scipy.special.comb(n0, 2.0)\nt3 = scipy.special.comb(t0, n2)\nt4 = t1 * t2\nanswer = t4 / t3\nprint(answer)" + ], + "Output Answer": [ + "0.5142857142857142" + ], + "split": "test" + }, + { + "Input": "if a coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land tails up on the first 4 flips and not on the last flip ?", + "Output Program": [ + "n0 = 1.0\nn1 = 2.0\nn2 = 5.0\nn3 = 2.0\nn4 = 3.0\nt0 = n1**min(n2, 5)\nanswer = 1 / t0\nprint(answer)" + ], + "Output Answer": [ + "0.03125" + ], + "split": "test" + } + ], + "Metadata": [ + { + "Answer": 0.16666666666666666, + "Options": "a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 3 , e ) \u00bd", + "Correct": "b", + "Explanation": "we are asked to find the probability of one particular pattern : wrrw . total # of ways a gardener can plant these four bushes is the # of permutations of 4 letters wwrr , out of which 2 w ' s and 2 r ' s are identical , so 4 ! / 2 ! 2 ! = 6 ; so p = 1 / 6 answer : b ." + }, + { + "Answer": 0.5, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these", + "Correct": "a", + "Explanation": "\"explanation : number greater than 3 is 4 , 5 , 6 , so only 3 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 3 / 6 = 1 / 2 answer : a\"" + }, + { + "Answer": 0.5, + "Options": "a ) 1 / 2 , b ) 1 / 5 , c ) 3 / 4 , d ) 1 / 3 , e ) 3 / 5", + "Correct": "a", + "Explanation": "this is a conditional probability question . basically we are told that out of 2 balls picked one is yellow and are asked to find the probability of the other one also to be yellow . so , possible options for this other ball would be : w , w , b , b ( as one yellow is already picked ) so the probability for that ball to be yellow too is 2 / 4 = 1 / 2 . option a" + }, + { + "Answer": 0.5357142857142857, + "Options": "a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14", + "Correct": "d", + "Explanation": "\"the required probability = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer .\"" + }, + { + "Answer": 0.25, + "Options": "a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 8 , d ) 1 / 16 , e ) 3 / 16", + "Correct": "b", + "Explanation": "the number of possible outcomes is 2 ^ 4 = 16 there are 4 ways to get exactly 3 heads . p ( exactly 3 heads ) = 4 / 16 = 1 / 4 the answer is b ." + }, + { + "Answer": 0.3076923076923077, + "Options": "a ) 8 / 13 , b ) 6 / 13 , c ) 3 / 13 , d ) 4 / 13 , e ) 1 / 13", + "Correct": "d", + "Explanation": "explanation : p ( s \u00e1 \u00b4 \u0153 k ) = p ( s ) + p ( k ) - p ( s \u00e2 \u02c6 \u00a9 k ) , where s denotes spade and k denotes king . p ( s \u00e1 \u00b4 \u0153 k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : d" + }, + { + "Answer": 0.015625, + "Options": "a ) 1 / 64 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 3", + "Correct": "a", + "Explanation": "\"solution - rolling dices is an independent event . the combinations to get 9 are ( 1,8 ) , ( 8,1 ) , ( 2,7 ) , ( 7,2 ) , ( 3,6 ) , ( 6,3 ) , ( 4,5 ) , ( 5,4 ) , and total combinations of both dices is 64 . the probability of getting 9 in first attempt is 8 / 64 = 1 / 8 . probability of getting 9 again in second attempt = ( 1 / 8 ) * ( 1 / 8 ) = 1 / 64 . ans a\"" + }, + { + "Answer": 0.23076923076923078, + "Options": "a ) 1 / 4 , b ) 1 / 12 , c ) 2 / 13 , d ) 3 / 13 , e ) 3 / 26", + "Correct": "d", + "Explanation": "the total number of cards is 52 . the number of jacks , queens , or kings is 12 . p ( jack , queen , king ) = 12 / 52 = 3 / 13 the answer is d ." + }, + { + "Answer": 6, + "Options": "a ) 12 , b ) 9 , c ) 8 , d ) 7 , e ) 6", + "Correct": "e", + "Explanation": "there are 18 balls in a jar . you take out 3 blue balls without putting 3 blue balls + 15 / 5 = 6 answer : e" + }, + { + "Answer": 25.00000000000001, + "Options": "a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 %", + "Correct": "b", + "Explanation": "\"could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . w = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %\"" + }, + { + "Answer": 0.07739938080495357, + "Options": "a ) 25 / 323 , b ) 21 / 969 , c ) 28 / 989 , d ) 74 / 879 , e ) 23 / 589", + "Correct": "a", + "Explanation": "\"sample space = no . of ways 3 balls were drawn from urn = 19 c 3 = 969 no . ways 2 blue balls and 1 red were drawn from bag = 6 c 2 * 5 c 1 = 75 probability = 75 / 969 = 25 / 323 ans - a\"" + }, + { + "Answer": 115.74074074074076, + "Options": "a ) 116 , b ) 146 , c ) 152 , d ) 162 , e ) 170", + "Correct": "a", + "Explanation": "\"the probability of survival for each of the first 3 months of life is 1 - 1 / 6 = 5 / 6 , so of 200 newborn 200 * 5 / 6 * 5 / 6 * 5 / 6 = ~ 116 is expected to survive . answer : a .\"" + }, + { + "Answer": 20, + "Options": "a ) 2 , b ) 3 , c ) 4 , d ) 20 , e ) 12", + "Correct": "d", + "Explanation": "\"10 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 10 x 2 = 20 d\"" + }, + { + "Answer": 0.006787330316742082, + "Options": "a ) 8 / 2652 , b ) 18 / 2652 , c ) 1 / 2652 , d ) 12 / 2652 , e ) 16 / 2652", + "Correct": "b", + "Explanation": "2 possible cases : king - 10 or 10 - king ( 4 kings and 4 10 ) . either way , the total probability = 2 ( king - 10 ) = 2 ( 4 / 52 * 4 / 51 ) = 18 / 2652 . b is the correct answer ." + }, + { + "Answer": 0.5, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 3 / 4 , e ) 2 / 3", + "Correct": "a", + "Explanation": "we can get 2 tails in 3 c 2 ways . we can get 3 tails in exactly one way . also , there are a total out 8 possible outcomes . thus , the required probability = ( 3 + 1 ) / 8 = 1 / 2 answer : a" + }, + { + "Answer": 0.5357142857142857, + "Options": "a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14", + "Correct": "d", + "Explanation": "\"the required probability x = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer .\"" + }, + { + "Answer": 0.5, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 2 / 5 , e ) 2 / 7", + "Correct": "a", + "Explanation": "\"s = { h , t } e = { t } p ( e ) = 1 / 2 answer is a\"" + }, + { + "Answer": 0.2, + "Options": "a ) 1 / 720 , b ) 1 / 80 , c ) 1 / 10 , d ) 1 / 9 , e ) 1 / 5", + "Correct": "e", + "Explanation": "in order to become a secretary the person can not be chosen as a president . this can be done in 9 / 10 ways . then to become a secretary this person ' s probability is 1 / 9 probability to become a secretary = ( 9 / 10 * 1 / 9 ) = 1 / 10 similar concept for treasurer . probability of not president = 9 / 10 , probability of not secretary = 8 / 9 , probability of treasurer 1 / 8 probability to become a treasurer = ( 9 / 10 * 8 / 9 * 1 / 8 ) = 1 / 10 since , the problem is saying parry secretary or treasurer it is addition : 1 / 10 + 1 / 10 = 1 / 5 = e" + }, + { + "Answer": 0.06666666666666667, + "Options": "a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 15", + "Correct": "e", + "Explanation": "\"total pairs = 16 c 2 = 120 ; same color pairs = 8 c 1 * 1 c 1 = 8 ; prob = 1 / 15 ans e\"" + }, + { + "Answer": 0.058823529411764705, + "Options": "a ) 2 / 5 , b ) 11 / 5 , c ) 1 / 9 , d ) 1 / 17 , e ) 1 / 5", + "Correct": "d", + "Explanation": "\"can be tackled in this way as well : probability of selecting any 1 out of 10 shoes = 18 / 18 = 1 probability of selecting the next shoe ( out of 18 available ) having the same color = 1 / 18 ( as after selecting the 1 st one , there is only 1 another shoe left with the same color ) . thus the total probability = 1 * 1 / 17 = 1 / 17 . d is the correct answer .\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 35 , b ) 1 / 10 , c ) 1 / 3 , d ) 1 / 4 , e ) 1 / 5", + "Correct": "c", + "Explanation": "\"first , total ways to select for all boys and girls , i . e 4 ! / ( 2 ! * 2 ! ) = 4 * 3 * 2 * 1 / 2 * 1 * 2 * 1 = 6 then there are one two way girls can have all same colors , either white or black . the number of ways in which 2 girls can select 2 white balls = 2 c 2 = 1 the number of ways in which 2 girls can select 2 black balls = 2 c 2 = 1 therefore , total favorable outcomes / total outcomes = 2 / 6 = 1 / 3 c\"" + }, + { + "Answer": 0.8, + "Options": "a ) . 5 , b ) 0.6 , c ) 0.4 , d ) 0.7 , e ) 0.8", + "Correct": "e", + "Explanation": "out of 10 minutes there is only 2 minutes chance of getting harbour line train and there is 8 minutes chance for getting main line train . so , probability = 8 / 10 = 0.8 answer : e" + }, + { + "Answer": 0.07692307692307693, + "Options": "a ) 1 / 13 , b ) 1 / 15 , c ) 11 / 9 , d ) 1 / 10 , e ) 1 / 25", + "Correct": "a", + "Explanation": "\"can be tackled in this way as well : probability of selecting any 1 out of 14 shoes = 14 / 14 = 1 probability of selecting the next shoe ( out of 14 available ) having the same color = 1 / 14 ( as after selecting the 1 st one , there is only 1 another shoe left with the same color ) . thus the total probability = 1 * 1 / 13 = 1 / 13 . a is the correct answer .\"" + }, + { + "Answer": 30, + "Options": "a ) a ) 30 , b ) b ) 45 , c ) c ) 50 , d ) d ) 55 , e ) e ) 60", + "Correct": "a", + "Explanation": "\"approach 1 at - least 1 paper back = total - no paper back 7 c 4 - 5 c 4 = 30 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 2 c 1 * 5 c 3 + 2 c 2 * 5 c 2 = 30 answer is a\"" + }, + { + "Answer": 0.13333333333333333, + "Options": "a ) 1 / 13 , b ) 1 / 14 , c ) 1 / 2 , d ) 2 / 15 , e ) 2 / 16", + "Correct": "d", + "Explanation": "\"taking 4 red shoe the probablity is 4 c 2 from 10 shoes probablity of taking 2 red shoe is 4 c 2 / 10 c 2 = 2 / 15 answer : d\"" + }, + { + "Answer": 0.47619047619047616, + "Options": "a ) 4 / 15 , b ) 6 / 17 , c ) 8 / 19 , d ) 10 / 21 , e ) 12 / 23", + "Correct": "d", + "Explanation": "\"the total number of ways to choose 6 children from 10 is 10 c 6 = 210 the number of ways to choose 3 boys and 3 girls is 5 c 3 * 5 c 3 = 10 * 10 = 100 p ( 3 boys and 3 girls ) = 100 / 210 = 10 / 21 the answer is d .\"" + }, + { + "Answer": 2, + "Options": "a ) 1 / 4 , b ) 1 / 2 , c ) 1 / 2 , d ) 5 / 8 , e ) 2", + "Correct": "e", + "Explanation": "\"my complicated version of your simple approach let the 4 black balls be bbbband 4 red balls be rrrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 8 ! ) / ( 4 ! x 4 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 7 slots and 3 black ( bbb ) and 4 red ( rrrr ) they can be arranged in ( 7 ! ) / ( 4 ! x 3 ! ) hence required probability = [ ( 8 ! ) / ( 4 ! x 4 ! ) ] / [ ( 7 ! ) / ( 4 ! x 3 ! ) ] = 2 e\"" + }, + { + "Answer": 0.42857142857142855, + "Options": "a ) 2 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 1 / 2", + "Correct": "d", + "Explanation": "p ( 1 st black , 2 nd white ) = 3 / 7 * 3 / 6 = 9 / 42 ; p ( 1 st white , 2 nd black ) = 3 / 7 * 3 / 6 = 9 / 42 . p = 9 / 42 + 9 / 42 = 18 / 42 = 3 / 7 . answer : d ." + }, + { + "Answer": 133.984375, + "Options": "a ) 134 , b ) 146 , c ) 152 , d ) 162 , e ) 170", + "Correct": "a", + "Explanation": "\"the probability of survival for each of the first 3 months of life is 1 - 1 / 8 = 7 / 8 , so of 200 newborn 200 * 7 / 8 * 7 / 8 * 7 / 8 = ~ 134 is expected to survive . answer : a .\"" + }, + { + "Answer": 71, + "Options": "a ) 65 , b ) 87 , c ) 71 , d ) 60 , e ) 66", + "Correct": "c", + "Explanation": "c 71 combined age in 15 years = 107 . 4 \u00d7 15 = 60 , therefore combined age now is 107 \u2013 60 = 47 . in 6 years time , therefore , combined age will be 47 + 24 ( 4 \u00d7 6 ) = 71 ." + }, + { + "Answer": 0.4, + "Options": "a ) 1 / 5 , b ) 2 / 5 , c ) 3 / 5 , d ) 4 / 5 , e ) 6 / 5", + "Correct": "b", + "Explanation": "total number of arrangement is = 5 ! = 120 favorable event i . e a and b can be arranged in 2 ! and the remaining can be arranged in 4 ! since ab can be in any of the four positions . so 4 ! * 2 ! / 5 ! = 2 / 5 answer : b" + }, + { + "Answer": 0.875, + "Options": "a ) 3 / 4 , b ) 5 / 8 , c ) 7 / 8 , d ) 9 / 16 , e ) 15 / 16", + "Correct": "c", + "Explanation": "\"p ( missing all 3 ) = ( 1 / 2 ) ^ 3 = 1 / 8 p ( success on at least one attempt ) = 1 - 1 / 8 = 7 / 8 the answer is c .\"" + }, + { + "Answer": 36, + "Options": "a ) 144 , b ) 288 , c ) 36 , d ) 256 , e ) none", + "Correct": "c", + "Explanation": "\"solution : let the arrangement be , b g b g b g b 34 boys can be seated in 3 ! ways . girl can be seated in 3 ! ways . required number of ways , = 3 ! * 3 ! = 36 . answer : option c\"" + }, + { + "Answer": 0.875, + "Options": "a ) 3 / 4 , b ) 1 / 4 , c ) 3 / 8 , d ) 7 / 8 , e ) none of these", + "Correct": "d", + "Explanation": "explanation : here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting at most two heads . then e = { ttt , tth , tht , htt , thh , hth , hht } . p ( e ) = n ( e ) / n ( s ) = 7 / 8 . answer is d" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6", + "Correct": "b", + "Explanation": "probably the best way to solve would be to use 1 - p ( opposite event ) = 1 - p ( odd ) = 1 - p ( odd ) * p ( odd ) = 1 - 2 / 4 * 2 / 3 = 8 / 12 = 1 / 3 . answer : b ." + }, + { + "Answer": 0.038461538461538464, + "Options": "a ) 1 / 13 , b ) 1 / 20 , c ) 1 / 26 , d ) 1 / 50 , e ) 1 / 100", + "Correct": "c", + "Explanation": "why do you calculate the total combinations and everything , it is much easier to calculate as follows : 3 / 13 * 1 * 1 * 1 * 1 . . . . . * 2 / 12 = 6 / 156 = 1 / 26 this is because 3 / 13 = probability that first letter will be e and 2 / 12 = probability that thefirstletter will be r . between everything else cuts down to 1 * 1 * 1 . . . e . g . probability q of the second letter is 11 / 11 = 1 , > we do not care what ' s in between . c" + }, + { + "Answer": 0.21428571428571427, + "Options": "a ) 1 / 3 , b ) 1 / 14 , c ) 3 / 14 , d ) 1 / 15 , e ) 1 / 16", + "Correct": "c", + "Explanation": "\"taking 2 red shoe the probability is 4 c 2 from 8 shoes probability of taking 2 red shoes is 4 c 2 / 8 c 2 = 3 / 14 answer : c\"" + }, + { + "Answer": 0.14814814814814814, + "Options": "a ) 2 / 27 , b ) 1 / 9 , c ) 1 / 3 , d ) 4 / 27 , e ) 2 / 9", + "Correct": "d", + "Explanation": "p = 2 \u2217 39 \u2217 29 = 427 p = 2 \u2217 39 \u2217 29 = 427 we are multiplying by 2 as there are two possible wining scenarios rw and wr . answer : d ." + }, + { + "Answer": 0.125, + "Options": "a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 3 , d ) 1 / 2 , e ) 3 / 4", + "Correct": "a", + "Explanation": "( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 8 answer : a" + }, + { + "Answer": 1.75, + "Options": "a ) 7 / 4 , b ) 1 / 2 , c ) 1 / 2 , d ) 5 / 8 , e ) 2 / 3", + "Correct": "a", + "Explanation": "my complicated version of your simple approach let the 5 black balls be bbbbb and 3 red balls be rrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 7 ! ) / ( 4 ! x 3 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 7 slots and 3 black ( bbb ) and 3 red ( rrr ) they can be arranged in ( 6 ! ) / ( 3 ! x 3 ! ) hence required probability = [ ( 7 ! ) / ( 4 ! x 3 ! ) ] / [ ( 6 ! ) / ( 3 ! x 3 ! ) ] ans = 7 / 4 a" + }, + { + "Answer": 0.75, + "Options": "a ) 3 / 4 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16", + "Correct": "a", + "Explanation": "\"bag consists of 20 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 20 - 5 - 9 = 6 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 20 + 6 / 20 = 15 / 20 = 3 / 4 hence , answer will be a .\"" + }, + { + "Answer": 1134, + "Options": "a ) 1100 , b ) 1200 , c ) 1235 , d ) 1354 , e ) 1140", + "Correct": "e", + "Explanation": "there 10 questions in part a out of which 8 question can be chosen as = 10 c 8 . similarly , 5 questions can be chosen from 10 questions of part b as = 10 c 5 . hence , total number of ways , = 10 c 8 * 10 c 5 = [ 10 ! / ( 2 ! 8 ! ) ] * [ 10 ! / ( 5 ! * 5 ) ] = { 10 * 9 / 2 } * { 10 * 9 * 8 * 7 * 6 / ( 5 * 4 * 3 * 2 * 1 ) } = 1140 . e" + }, + { + "Answer": 25, + "Options": "a ) 18 , b ) 21 , c ) 25 , d ) 30 , e ) 35", + "Correct": "c", + "Explanation": "the total number of ways to choose 4 managers is 7 c 4 = 35 we need to subtract the number of groups which include the two managers , which is 5 c 2 = 10 . 35 - 10 = 25 the answer is c ." + }, + { + "Answer": 24, + "Options": "a ) 144 , b ) 288 , c ) 24 , d ) 256 , e ) none", + "Correct": "c", + "Explanation": "solution : let the arrangement be , b b b b g 4 boys can be seated in 4 ! ways . girl can be seated in 1 ! ways . required number of ways , = 4 ! * 1 ! = 24 . answer : option c" + }, + { + "Answer": 2880, + "Options": "a ) 144 , b ) 288 , c ) 12 , d ) 256 , e ) 2880", + "Correct": "e", + "Explanation": "\"solution : let the arrangement be , g b g b g b g b g 4 boys can be seated in 4 ! ways . girl can be seated in 3 ! ways . required number of ways , = 4 ! * 5 ! = 2880 . answer : option e\"" + }, + { + "Answer": 0.875, + "Options": "a ) 3 / 4 , b ) 1 / 4 , c ) 7 / 8 , d ) 1 / 8 , e ) 6 / 8", + "Correct": "c", + "Explanation": "using 0 and 1 for head and tail respectively and considering binary system , we find that out of 16 possibilities , 14 are desired options . then probability = 14 / 16 = 7 / 8 answer : c" + }, + { + "Answer": 0.625, + "Options": "a ) 3 / 5 , b ) 5 / 8 , c ) 6 / 7 , d ) 7 / 8 , e ) 6 / 7", + "Correct": "b", + "Explanation": "\"since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . \u2234 required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b\"" + }, + { + "Answer": 0.6, + "Options": "a ) 3 / 5 , b ) 1 / 5 , c ) 2 / 5 , d ) 4 / 5 , e ) 1", + "Correct": "a", + "Explanation": "using the first example , here is the probability of that exact sequence occurring : wwgg = ( 3 / 6 ) ( 2 / 5 ) ( 3 / 4 ) ( 2 / 3 ) = 36 / 360 = 1 / 10 each of the other 5 options will yield the exact same probability . . . . wgwg = ( 3 / 6 ) ( 3 / 5 ) ( 2 / 4 ) ( 2 / 3 ) = 36 / 360 = 1 / 10 so we have 6 different options that each produce a 1 / 10 chance of occurring . 6 ( 1 / 10 ) = 6 / 10 = 3 / 5 final answer : a" + }, + { + "Answer": 0.9166666666666666, + "Options": "a ) 11 / 12 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16", + "Correct": "a", + "Explanation": "bag consists of 60 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 60 - 5 - 9 = 46 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 60 + 46 / 60 = 55 / 60 = 11 / 12 hence , answer will be a ." + }, + { + "Answer": 55, + "Options": "a ) 48 , b ) 55 , c ) 58 , d ) 62 , e ) 70", + "Correct": "b", + "Explanation": "\"the total number of ways to choose 4 managers is 8 c 4 = 70 we need to subtract the number of groups which include the two managers , which is 6 c 2 = 15 . 70 - 15 = 55 the answer is b .\"" + }, + { + "Answer": 0.125, + "Options": "a ) 1 / 4 , b ) 1 / 6 , c ) 1 / 8 , d ) 1 / 12 , e ) 1 / 16", + "Correct": "c", + "Explanation": "on the first toss , the coin will land on one side or the other . on the second toss , the probability of landing on the same side is 1 / 2 . on the third toss , the probability of landing on the same side is 1 / 2 . on the fourth toss , the probability of landing on the same side is 1 / 2 . p ( same side all four times ) = 1 / 2 * 1 / 2 * 1 / 2 = 1 / 8 . the answer is c ." + }, + { + "Answer": 0.10526315789473684, + "Options": "a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10", + "Correct": "b", + "Explanation": "\"method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability q = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b\"" + }, + { + "Answer": 0.3, + "Options": "a ) 10 / 50 , b ) 15 / 50 , c ) 8 / 50 , d ) 3 / 50 , e ) 7 / 50", + "Correct": "b", + "Explanation": "let x be the event of selecting a prime number . x = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 } n ( x ) = 15 , n ( s ) = 50 hence , the required probability is 15 / 50 . answer : b" + }, + { + "Answer": 0.9375, + "Options": "a ) 3 / 4 , b ) 5 / 8 , c ) 7 / 8 , d ) 11 / 16 , e ) 15 / 16", + "Correct": "e", + "Explanation": "p ( missing all 4 ) = ( 1 / 2 ) ^ 4 = 1 / 16 p ( success on at least one attempt ) = 1 - 1 / 16 = 15 / 16 the answer is e ." + }, + { + "Answer": 0.08333333333333333, + "Options": "a ) 1 / 12 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 2 , e ) 1 / 6", + "Correct": "a", + "Explanation": "probability of getting mcq answer correct = 1 / 3 probability of getting one true / false answer correct = 1 / 2 probability of getting all 3 answers correct = 1 / 3 * 1 / 2 * 1 / 2 = 1 / 12 answer : a" + }, + { + "Answer": 144, + "Options": "a ) 120 , b ) 144 , c ) 160 , d ) 210 , e ) 180", + "Correct": "b", + "Explanation": "\"4 boys can be seated in 4 ! three girls can be seated in 3 ! required number = 4 ! 3 ! = 144 answer is b\"" + }, + { + "Answer": 0.1388888888888889, + "Options": "a ) 3 / 13 , b ) 2 / 31 , c ) 5 / 36 , d ) 7 / 41 , e ) 7 / 43", + "Correct": "c", + "Explanation": "s = { ( 1,1 ) , ( 1,2 ) , ( 1,3 ) , ( 1,4 ) , ( 1,5 ) , ( 1,6 ) , ( 4,1 ) , ( 4,2 ) , ( 4,3 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 2,1 ) , ( 2,2 ) , ( 2,3 ) , ( 2,4 ) , ( 2,5 ) , ( 2,6 ) , ( 5,1 ) , ( 5,2 ) , ( 5,3 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 3,1 ) , ( 3,2 ) , ( 3,3 ) , ( 3,4 ) , ( 3,5 ) , ( 3,6 ) , ( 6,1 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) } n ( s ) = 36 set of total of 6 , a = { ( 1,5 ) , ( 5,1 ) , ( 2,4 ) , ( 4,2 ) , ( 3,3 ) } ; n ( a ) = 5 p ( total of 6 ) = n ( a ) n ( s ) n ( a ) n ( s ) = 5 / 36 c" + }, + { + "Answer": 0.0625, + "Options": "a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 12 , d ) 1 / 16 , e ) 1 / 32", + "Correct": "d", + "Explanation": "\"on the first toss , the coin will land on one side or the other . on the second toss , the probability of landing on the same side is 1 / 2 . on the third toss , the probability of landing on the same side is 1 / 2 . on the fourth toss , the probability of landing on the same side is 1 / 2 . on the fifth toss , the probability of landing on the same side is 1 / 2 . p ( same side all five times ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 . the answer is d .\"" + }, + { + "Answer": 0.84, + "Options": "a ) 3 / 5 , b ) 4 / 5 , c ) 17 / 25 , d ) 21 / 25 , e ) 43 / 50", + "Correct": "d", + "Explanation": "there can be an odd product only if both integers are odd . p ( odd product ) = 4 / 10 * 2 / 5 = 4 / 25 p ( even product ) = 1 - 4 / 25 = 21 / 25 the answer is d ." + }, + { + "Answer": 40, + "Options": "a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 %", + "Correct": "d", + "Explanation": "\"total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability w = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d\"" + }, + { + "Answer": 2, + "Options": "a ) 1 / 4 , b ) 1 / 2 , c ) 1 / 2 , d ) 2 , e ) 2 / 3", + "Correct": "d", + "Explanation": "\"my complicated version of your simple approach let the 3 black balls be bbb and 3 red balls be rrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 6 ! ) / ( 3 ! x 3 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 5 slots and 2 black ( bb ) and 3 red ( rrr ) they can be arranged in ( 5 ! ) / ( 2 ! x 3 ! ) hence required probability = [ ( 6 ! ) / ( 3 ! x 3 ! ) ] / [ ( 5 ! ) / ( 2 ! x 3 ! ) ] = 2 d\"" + }, + { + "Answer": 0.12745098039215685, + "Options": "a ) 102 / 102 , b ) 5 / 102 , c ) 1 / 102 , d ) 8 / 102 , e ) 13 / 102", + "Correct": "e", + "Explanation": "solution let s be the sample space . then , n ( s ) = 52 c 2 = ( 52 \u00d7 51 ) / ( 2 \u00d7 1 ) = 1326 . let e = event of getting 1 spade and 1 heart . n ( e ) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = ( 13 c 1 * 13 c 1 ) = 169 . p ( e ) = n ( e ) / n ( s ) = 169 / 1326 = 13 / 102 . answer e" + }, + { + "Answer": 0.44367283950617287, + "Options": "a ) 15 / 36 , b ) 175 / 432 , c ) 575 / 1296 , d ) 583 / 1296 , e ) 1 / 2", + "Correct": "c", + "Explanation": "there are three possible outcomes in the game : 1 . b wins 2 . j wins 3 . draw so , 1 = p ( j ) + p ( b ) + p ( draw ) 1 = p ( j ) + p ( b ) + p ( draw ) the probability of b winning or j winning the game is same . so , 1 = 2 \u2217 p ( j ) + p ( draw ) 1 = 2 \u2217 p ( j ) + p ( draw ) so , we need to calculate the probability of a draw . a draw can happen when both will have the same sum . the possible sums are - 2 , 34 , 56 , 78 , 910 , 1112 prob . of each sum is - 2 - 11 - 1 / 36 3 - 1 2 , 2 1 - 2 / 36 4 - 1 3 , 2 2 , 3 1 - 3 / 36 5 - 1 4 , 2 3 , 3 2 , 41 - 4 / 36 6 - 1 5 , 2 4 , 3 3 , 4 2 , 5 1 - 5 / 36 7 - 1 6 , 2 5 , 3 4 , 4 3 , 5 2 , 61 - 6 / 36 8 - 2 6 , 3 5 , 4 4 , 5 3 , 6 2 , - 5 / 36 9 - 3 6 , 4 5 , 5 4 , 63 - 4 / 36 10 - 4 6 , 5 5 , 6 4 - 3 / 36 11 - 5 6 , 6 5 - 2 / 36 12 - 6 6 - 1 / 36 now , since both have to get the same sum , the probability will be same for both and the combined probability will be the square of the probabilities i . e . 2 - ( 1 / 36 ) ^ 2 3 - ( 2 / 36 ) ^ 2 etc . squaring the probabilities and adding them will give 146 / 1296 so , p ( d ) = 146 / 1296 1 = 2 * p ( j ) + p ( d ) 2 * p ( j ) = 1150 / 1296 p ( j ) = 575 / 1296 ans . c" + }, + { + "Answer": 0.5357142857142857, + "Options": "a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14", + "Correct": "d", + "Explanation": "the required probability q = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer ." + }, + { + "Answer": 240, + "Options": "a ) a . 240 , b ) b . 808250 , c ) c . 502450 , d ) d . 784200 , e ) e . 302400", + "Correct": "a", + "Explanation": "ways for cages that are too small = 5 * 4 * 3 * 2 = 120 since we have 2 cages and 2 animal left therefore ways for 2 cages = 2 ! total ways = 120 * 2 ! = 604800 answer a" + }, + { + "Answer": 0.08333333333333333, + "Options": "a ) 1 / 16 , b ) 2 / 14 , c ) 1 / 12 , d ) 1 / 19 , e ) 1 / 17", + "Correct": "c", + "Explanation": "c 1 / 12 all possible cases can be 36 ( 6 * 6 ) case we need : [ ( 4,6 ) , ( 5,5 ) , ( 6,4 ) ] = 3 probability = > 3 / 36 = 1 / 12" + }, + { + "Answer": 0.368421052631579, + "Options": "a ) 7 / 19 , b ) 6 / 19 , c ) 5 / 19 , d ) 4 / 19 , e ) none of these", + "Correct": "a", + "Explanation": "\"explanation : please remember that maximum portability is 1 . so we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs . so here we go , total cases of non defective bulbs 16 c 2 = 16 \u2217 15 / 2 \u2217 1 = 120 total cases = 20 c 2 = 20 \u2217 19 / 2 \u2217 1 = 190 probability = 120 / 190 = 12 / 19 p of at least one defective = 1 \u2212 12 / 19 = 7 / 19 option a\"" + }, + { + "Answer": 0.7142857142857143, + "Options": "a ) 2 / 3 , b ) 6 / 7 , c ) 5 / 7 , d ) 8 / 7 , e ) 4 / 3", + "Correct": "c", + "Explanation": "total draws = prizes + blanks = 10 + 25 = 35 probability of getting a blank = 25 / 35 = 5 / 7 correct option is c" + }, + { + "Answer": 0.9642857142857143, + "Options": "a ) 25 / 28 , b ) 27 / 28 , c ) 30 / 28 , d ) 40 / 28 , e ) 50 / 28", + "Correct": "b", + "Explanation": "there are 6 consonants and 2 vowels in khantkar . probability that at least one of the selected letters is a consonant = 1 - ( probability of both vowels ) probability of both vowels = 2 c 2 / 8 c 2 = 1 / 28 so , answer is 1 - 1 / 28 = 27 / 28 answer : b" + }, + { + "Answer": 0.4995, + "Options": "a ) 0.3995 , b ) 0.4995 , c ) 0.5995 , d ) 0.6995 , e ) 0.7995", + "Correct": "b", + "Explanation": "probability of a choosing 1 and b greater = ( 1 / 1000 ) * ( 999 / 1000 ) ; probability of a choosing 2 and b greater = ( 1 / 1000 ) * ( 998 / 1000 ) ; probability of a choosing 3 and b greater = ( 1 / 1000 ) * ( 997 / 1000 ) ; . . . ; probability of a choosing 998 and b greater = ( 1 / 1000 ) * ( 2 / 1000 ) ; probability of a choosing 999 and b greater = ( 1 / 1000 ) * ( 1 / 1000 ) ; total = ( 1 / 1000 ) * ( 1 / 1000 ) * ( 999 + 998 + 997 + . . . + 2 + 1 ) = ( 1 / 1000 ) * ( 1 / 1000 ) * ( 999 * 1000 / 2 ) = 0.4995 answer : b" + }, + { + "Answer": 0.20000000000000007, + "Options": "a ) 1 / 720 , b ) 1 / 80 , c ) 1 / 10 , d ) 1 / 9 , e ) 1 / 5", + "Correct": "e", + "Explanation": "in order to become a secretary the person can not be chosen as a president . this can be done in 9 / 10 ways . then to become a secretary this person ' s probability is 1 / 9 probability to become a secretary = ( 9 / 10 * 1 / 9 ) = 1 / 10 similar concept for treasurer . probability of not president = 9 / 10 , probability of not secretary = 8 / 9 , probability of treasurer 1 / 8 probability to become a treasurer = ( 9 / 10 * 8 / 9 * 1 / 8 ) = 1 / 10 since , the problem is saying jarry secretary or treasurer it is addition : 1 / 10 + 1 / 10 = 1 / 5 = e" + }, + { + "Answer": 0.25, + "Options": "a ) 1 / 2 , b ) 1 / 4 , c ) 2 / 3 , d ) 2 / 5 , e ) 3 / 7", + "Correct": "b", + "Explanation": "total number of cases = 6 * 6 = 36 favourable cases = [ ( 1,3 ) , ( 1,5 ) , ( 1,1 ) , ( 3,1 ) , ( 3,3 ) , ( 3,5 ) , ( 5,1 ) , ( 5,3 ) , ( 5,5 ) ] = 9 so probability = 9 / 36 = 1 / 4 answer is b" + }, + { + "Answer": 0.7391304347826086, + "Options": "a ) 13 / 23 , b ) 16 / 23 , c ) 11 / 23 , d ) 17 / 23 , e ) 18 / 23", + "Correct": "d", + "Explanation": "total no . of ways of selecting 4 consecutive letters = > 23 [ like abcd , bcde , efgh , . . etc ] no . of ways of having one is a vowel = > abcd , bcde , cdef , defg , efgh , fghi , ghij , hijk , ijkl , lmno , mnop , nopq , opqr , rstu , stuv , tuvw , uvwx . . . = 17 p ( one of letter is vowel ) = 17 / 23 answer : d" + }, + { + "Answer": 0.08888888888888889, + "Options": "a ) 2 / 25 , b ) 4 / 75 , c ) 7 / 75 , d ) 3 / 5 , e ) 4 / 45", + "Correct": "e", + "Explanation": "probability of sand dune blown out = 2 / 3 probability that it has treasure = 1 / 5 probability that it has lucky coupon = 2 / 3 total probability = 2 / 3 * 1 / 5 * 2 / 3 = 4 / 45 answer : e" + }, + { + "Answer": 510.3, + "Options": "a ) 511 , b ) 546 , c ) 552 , d ) 562 , e ) 570", + "Correct": "a", + "Explanation": "\"number of newborns that can die in first month = 1 / 10 * 700 = 70 survived = 630 number of newborns that can die in second month = 1 / 10 * 630 = 63 survived = 567 number of newborns that can die in third month = 1 / 10 * 567 = 56 survived = 511 answer : a\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 2 / 3 , b ) 3 / 5 , c ) 4 / 7 , d ) 8 / 9 , e ) 1 / 3", + "Correct": "e", + "Explanation": "\"e = { 5,6 } n ( e ) = 2 p ( s ) = n ( e ) / n ( s ) = 2 / 6 = 1 / 3 ans : e\"" + }, + { + "Answer": 0.10526315789473684, + "Options": "a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10", + "Correct": "b", + "Explanation": "\"method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability y = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b\"" + }, + { + "Answer": 0.234375, + "Options": "a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64", + "Correct": "a", + "Explanation": "\"chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days a = 15 / 64 ans a it is .\"" + }, + { + "Answer": 30, + "Options": "a ) 2 , b ) 3 , c ) 30 , d ) 6 , e ) 12", + "Correct": "c", + "Explanation": "15 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 15 x 2 = 30 c" + }, + { + "Answer": 71, + "Options": "a ) 61 , b ) 78 , c ) 71 , d ) 68 , e ) 94", + "Correct": "c", + "Explanation": "c 71 combined age in 15 years = 107 . 4 \u00d7 15 = 60 , therefore combined age now is 107 \u2013 60 = 47 . in 6 years time , therefore , combined age will be 47 + 24 ( 4 x 6 ) = 71 ." + }, + { + "Answer": 0.02040816326530612, + "Options": "a ) 1 / 144 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 49", + "Correct": "e", + "Explanation": "\"solution - rolling dices is an independent event . the combinations to get 8 are ( 1,7 ) , ( 7,1 ) , ( 2,6 ) , ( 6,2 ) , ( 3,5 ) , ( 5,3 ) , ( 4 , , 4 ) and total combinations of both dices is 49 . the probability of getting 8 in first attempt is 7 / 49 = 1 / 7 . probability of getting 8 again in second attempt = ( 1 / 7 ) * ( 1 / 7 ) = 1 / 49 . ans e\"" + }, + { + "Answer": 182.25, + "Options": "a ) 140 , b ) 146 , c ) 152 , d ) 162 , e ) 183", + "Correct": "e", + "Explanation": "\"the probability of survival for each of the first 3 months of life is 1 - 1 / 10 = 9 / 10 , so of 250 newborn 250 * 9 / 10 * 9 / 10 * 9 / 10 = ~ 183 is expected to survive . answer : e .\"" + }, + { + "Answer": 220, + "Options": "a ) 220 . , b ) 240 , c ) 260 , d ) 280 , e ) 300", + "Correct": "a", + "Explanation": "we have 11 english and 9 french books , no french books should be adjacent . imagine 11 english books in a row and empty slots like below : * e * e * e * e * e * e * e * e * e * e * e * now if 9 french books would be placed in 12 empty slots , all french books will be separated by english books . so we can ` ` choose ' ' 9 empty slots from 12 available for french books , which is 12 c 9 = 220 . ans : a _________________" + }, + { + "Answer": 40, + "Options": "a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 %", + "Correct": "d", + "Explanation": "\"total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability r = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d\"" + }, + { + "Answer": 0.03125, + "Options": "a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 8 , d ) 1 / 16 , e ) 1 / 32", + "Correct": "e", + "Explanation": "\"p ( hhttt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 the answer is e .\"" + }, + { + "Answer": 0.4666666666666667, + "Options": "a ) 1 / 13 , b ) 1 / 14 , c ) 1 / 15 , d ) 7 / 15 , e ) 7 / 16", + "Correct": "d", + "Explanation": "\"taking 2 red shoe the probablity is 7 c 2 from 10 shoes probablity of taking 2 red shoe is 7 c 2 / 10 c 2 = 7 / 15 answer : d\"" + }, + { + "Answer": 25.00000000000001, + "Options": "a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 %", + "Correct": "b", + "Explanation": "\"could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . q = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %\"" + }, + { + "Answer": 218.70000000000002, + "Options": "a ) 219 , b ) 246 , c ) 252 , d ) 262 , e ) 270", + "Correct": "a", + "Explanation": "\"number of newborns that can die in first month = 1 / 10 * 300 = 30 survived = 270 number of newborns that can die in second month = 1 / 10 * 270 = 27 survived = 243 number of newborns that can die in third month = 1 / 10 * 243 = 24 survived = 219 answer : a\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6", + "Correct": "b", + "Explanation": "\"probably the best way to solve would be to use 1 - p ( opposite event ) = 1 - p ( odd ) = 1 - p ( odd ) * p ( odd ) = 1 - 2 / 4 * 2 / 3 = 8 / 12 = 1 / 3 . answer : b .\"" + }, + { + "Answer": 1.4, + "Options": "a ) 1 / 4 , b ) 7 / 5 , c ) 1 / 2 , d ) 5 / 8 , e ) 2 / 3", + "Correct": "b", + "Explanation": "\"my complicated version of your simple approach let the 5 black balls be bbbbb and 2 red balls be rrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 7 ! ) / ( 5 ! x 2 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 6 slots and 4 black ( bbbb ) and 2 red ( rrr ) they can be arranged in ( 7 ! ) / ( 4 ! x 2 ! ) hence required probability = [ ( 7 ! ) / ( 5 ! x 2 ! ) ] / [ ( 6 ! ) / ( 4 ! x 2 ! ) ] = 7 / 5 b\"" + }, + { + "Answer": 56, + "Options": "a ) 53 , b ) 54 , c ) 55 , d ) 56 , e ) 57", + "Correct": "d", + "Explanation": "\"first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 8 and for second = 7 so total methods = 8 * 7 = 56 answer : d\"" + }, + { + "Answer": 0.5, + "Options": "a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 4 , d ) 1 / 8 , e ) 3 / 8", + "Correct": "a", + "Explanation": "\"n ( s ) = 2 ^ 3 = 8 let e is the event of getting at least 2 tails n ( e ) = ttt , tth , htt , tht = 4 p ( e ) = n ( e ) / n ( s ) = 4 / 8 = 1 / 2 ans - a\"" + }, + { + "Answer": 215, + "Options": "a ) 216 , b ) 243 , c ) 215 , d ) 729 , e ) 728", + "Correct": "c", + "Explanation": "since each ring consists of six different letters , the total number of attempts possible with the three rings is = 6 * 6 * 6 = 216 . of these attempts , one of them is a successful attempt . maximum number of unsuccessful attempts = 216 - 1 = 215 . answer : c" + }, + { + "Answer": 11, + "Options": "a ) 6 , b ) 8 , c ) 10 , d ) 11 , e ) 15", + "Correct": "d", + "Explanation": "\"the total number of ways to choose 5 managers is 7 c 5 = 21 we need to subtract the number of groups which include the two managers , which is 5 c 3 = 10 . 21 - 10 = 11 the answer is d .\"" + }, + { + "Answer": 0.6666666666666666, + "Options": "a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1", + "Correct": "d", + "Explanation": "\"answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability g is 2 / 3 . d\"" + }, + { + "Answer": 0.68, + "Options": "a ) 17 / 65 , b ) 17 / 25 , c ) 17 / 28 , d ) 17 / 65 , e ) 17 / 39", + "Correct": "b", + "Explanation": "\"the number of exhaustive events = \u2075 \u2070 c \u2081 = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . required probability = 34 / 50 = 17 / 25 . answer : b\"" + }, + { + "Answer": 0.9937037037037038, + "Options": "a ) 17 / 2900 , b ) 2683 / 2700 , c ) 2496 / 2700 , d ) 2369 / 2700 , e ) 198 / 270", + "Correct": "b", + "Explanation": "in how many ways we can select a girl and a boy from 45 lady and 60 gentlemen ? in 45 * 60 = 2,700 ways . we have a total of 17 couples so , the probability of selecting a couple is 17 / 2,700 = 17 / 2700 . probability of there will not be a couple = 1 - ( 17 / 2700 ) = 2683 / 2700 ans - b" + }, + { + "Answer": 0.07142857142857142, + "Options": "a ) 1 / 7 , b ) 1 / 12 , c ) 1 / 14 , d ) 1 / 18 , e ) 1 / 21", + "Correct": "c", + "Explanation": "the total number of ways to sell 6 bottles from 10 is 10 c 6 = 210 . the number of ways to sell 2 bottles of apple juice is 6 c 2 * 4 c 4 = 15 * 1 = 15 p ( selling 2 bottles of apple juice ) = 15 / 210 = 5 / 70 = 1 / 14 the answer is c ." + }, + { + "Answer": 0.2222222222222222, + "Options": "a ) 1 / 4 , b ) 1 / 9 , c ) 4 / 3 , d ) 3 / 9 , e ) 2 / 9", + "Correct": "e", + "Explanation": "total number of outcomes possible = 36 probability of getting sum of 7 = 6 / 36 probability of getting sum of 11 = 2 / 36 probability that the sum is either 7 or 11 = 6 / 36 + 2 / 36 = 2 / 9 answer : e" + }, + { + "Answer": 40, + "Options": "a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 %", + "Correct": "d", + "Explanation": "\"total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability q = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d\"" + }, + { + "Answer": 1, + "Options": "a ) 1 , b ) 5 , c ) 8 , d ) 4 , e ) 3", + "Correct": "a", + "Explanation": "method - 1 gk will be even when 1 ) g is even and k is odd , probability of g even is ( 1 / 4 ) and probability of k odd is ( 1 / 3 ) , so probability of case ( 1 ) = ( 1 / 4 ) * ( 1 / 3 ) = ( 1 / 12 ) 2 ) g is odd and k is even , probability of g odd is ( 3 / 4 ) and probability of k even is ( 2 / 3 ) , so probability of case ( 2 ) = ( 3 / 4 ) * ( 2 / 3 ) = ( 6 / 12 ) 3 ) g is even and k is even , probability of g even is ( 1 / 4 ) and probability of k even is ( 2 / 3 ) , so probability of case ( 1 ) = ( 1 / 4 ) * ( 2 / 3 ) = ( 2 / 12 ) total favorable probability = ( 1 / 12 ) + ( 6 / 12 ) + ( 2 / 12 ) = ( 12 / 12 ) = 1 answer : option a" + }, + { + "Answer": 364.5, + "Options": "a ) 340 , b ) 346 , c ) 352 , d ) 365 , e ) 370", + "Correct": "d", + "Explanation": "number of newborns that can die in first month = 1 / 10 * 500 = 50 survived = 450 number of newborns that can die in second month = 1 / 10 * 450 = 45 survived = 405 number of newborns that can die in third month = 1 / 10 * 405 = 40 survived = 365 answer : d" + }, + { + "Answer": 0.1111111111111111, + "Options": "a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 25", + "Correct": "c", + "Explanation": "\"total pairs = 10 c 2 = 45 ; same color pairs = 5 c 1 * 1 c 1 = 5 ; prob = 1 / 9 or 2 / 10 * 1 / 9 * 5 = 1 / 9 ans c\"" + }, + { + "Answer": 0.2857142857142857, + "Options": "a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 2 / 7", + "Correct": "e", + "Explanation": "there are 366 days in a leap year : 52 weeks and 2 more days . so , 52 mondays and 2 days . these 2 days can be : { mon , tue } , { tue , wed } , { wed , thu } , { thu , fri } , { fri , sat } , { sat , sun } and { sun , mon } ( 7 cases ) . in order to have 53 mondays we should have either { mon , tuesday } or { sun , mon } case . the probability of that is 2 / 7 . answer : e ." + }, + { + "Answer": 0.6666666666666666, + "Options": "a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1", + "Correct": "d", + "Explanation": "answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability l is 2 / 3 . d" + }, + { + "Answer": 40, + "Options": "a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 %", + "Correct": "d", + "Explanation": "\"total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability e = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d\"" + }, + { + "Answer": 50, + "Options": "a ) 25.0 % , b ) 50.0 % , c ) 62.5 % , d ) 75.0 % , e ) 80.0 %", + "Correct": "b", + "Explanation": "\"p ( chile and not madagascar ) = 0.5 * 0.5 = 0.25 p ( madagascar and not chile ) = 0.5 * 0.5 = 0.25 total probability = 0.25 + 0.25 = 0.5 = 50 % the answer is b .\"" + }, + { + "Answer": 1540, + "Options": "a ) 3990 , b ) 1540 , c ) 1995 , d ) 3672 , e ) none of these", + "Correct": "b", + "Explanation": "\"in order that two books on french are never together , we must place all books as under x e x e x e x e x e . . . . . x e x where e denotes the position of an english book and x that of a french book . since there are 21 books on english , the number of places marked x are therefore , 22 . now , 19 places out of 22 can be chosen in 22 c 19 = 22 c 3 = ( 22 x 21 x 20 ) / ( 3 x 2 x 1 ) = 1540 ways answer b .\"" + }, + { + "Answer": 0.3626373626373626, + "Options": "a ) 12 / 91 , b ) 3 / 15 , c ) 12 / 15 , d ) 33 / 91 , e ) 3 / 91", + "Correct": "d", + "Explanation": "since jake owns 3 of the boots , the subset from which the 4 boots should be chosen are the 12 boots not owned by jake from the universe of 15 . the first boot can be one of the 12 from the 15 with probability 12 / 15 . the second boot can be one of the 11 from the 14 remaining with probability 11 / 14 . the third boot can be one of the 10 from the 13 remaining with probability 10 / 13 . the fourth boot can be one of the 9 from the 12 remaining with probability 9 / 12 . the total probability will be 12 / 15 \u00d7 11 / 14 \u00d7 10 / 13 \u00d7 9 / 12 . on cancellation , this comes to 33 / 91 . thus , the answer is d - 33 / 91" + }, + { + "Answer": 0.25, + "Options": "a ) 0.75 , b ) 1 , c ) 0.25 , d ) 0.5 , e ) 0.58", + "Correct": "c", + "Explanation": "radius of inner dartboard / overall radius of dartboard ( ( r / 2 ) ^ 2 ) / ( r ^ 2 ) = 1 / 4 = 0.25 answer : c" + }, + { + "Answer": 0.2857142857142857, + "Options": "a ) 1 / 7 , b ) 2 / 7 , c ) 3 / 7 , d ) 4 / 7 , e ) 1", + "Correct": "b", + "Explanation": "total number of outcomes possible , n ( s ) = 10 + 25 = 35 . total number of prize , n ( e ) = 10 p ( e ) = n ( e ) / n ( s ) = 10 / 35 = 2 / 7 answer is option b" + }, + { + "Answer": 0.03333333333333333, + "Options": "a ) 1 / 18 , b ) 1 / 24 , c ) 1 / 30 , d ) 1 / 36 , e ) 1 / 48", + "Correct": "c", + "Explanation": "the number of ways of choosing 3 balls from the jar is 10 c 3 = 120 . the number of ways of choosing 3 blue balls is 4 c 3 = 4 . p ( 3 blue balls ) = 4 / 120 = 1 / 30 . the answer is c ." + }, + { + "Answer": 0.68, + "Options": "a ) 17 / 25 , b ) 17 / 27 , c ) 17 / 18 , d ) 17 / 22 , e ) 17 / 09", + "Correct": "a", + "Explanation": "\"the number of exhaustive events = \u2075 \u2070 c \u2081 = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . required probability = 34 / 50 = 17 / 25 . answer : a\"" + }, + { + "Answer": 246, + "Options": "a ) 75 , b ) 120 , c ) 210 , d ) 246 , e ) 252", + "Correct": "d", + "Explanation": "\"paperbacks - 4 , hardbacks - 6 5 books in total and at least 1 from each . total combinations for 5 books = ( 1 pb , 4 hb ) + ( 4 pb , 1 hb ) + ( 3 pb , 2 hb ) + ( 2 pb , 3 hb ) 1 pb , 4 hb = 4 c 1 * 6 c 4 = 60 4 pb , 1 hb = 4 c 4 * 6 c 1 = 6 3 pb , 2 hb = 4 c 3 * 6 c 2 = 60 2 pb , 3 hb = 4 c 2 * 6 c 3 = 120 total combinations of 5 books = 60 + 6 + 60 + 120 = > 246 ans d .\"" + }, + { + "Answer": 0.1111111111111111, + "Options": "a ) 1 / 144 , b ) 1 / 36 , c ) 1 / 9 , d ) 1 / 6 , e ) 1 / 3", + "Correct": "c", + "Explanation": "\"solution - rolling dices is an independent event . the combinations to get 4 are ( 1,3 ) , ( 3,1 ) , ( 2,2 ) , and total combinations of both dices is 9 . the probability of getting 4 in first attempt is 3 / 9 = 1 / 3 . probability of getting 4 again in second attempt = ( 1 / 3 ) * ( 1 / 3 ) = 1 / 9 . ans c\"" + }, + { + "Answer": 0.6666666666666666, + "Options": "a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1", + "Correct": "d", + "Explanation": "\"answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability z is 2 / 3 . d\"" + }, + { + "Answer": 0.6588235294117647, + "Options": "a ) 1 / 33 , b ) 2 / 85 , c ) 1 / 3 , d ) 56 / 85 , e ) 11 / 85", + "Correct": "d", + "Explanation": "if we are to select 4 people from 9 couples without any restriction , how many ways can we make the selection ? 18 ! / 4 ! 14 ! = 3060 if we are to select 4 people from 9 couples with restriction that no married couple can both make it to the group , only a representative ? 9 ! / 4 ! 5 ! = 126 but we know that to select a person from each couple , take 2 possibilities 126 * 2 * 2 * 2 * 2 = 2016 probability = desired / all possibilities = 2016 / 3060 = 56 / 85 answer : d" + }, + { + "Answer": 0.6, + "Options": "a ) 1 / 10 , b ) 4 / 9 , c ) 1 / 2 , d ) 9 / 15 , e ) 2 / 3", + "Correct": "d", + "Explanation": "\"the total number of ways to choose 2 children from 6 is 6 c 2 = 15 the number of ways to choose 1 boy and 1 girl is 3 * 3 = 9 p ( 1 boy and 1 girl ) = 9 / 15 the answer is d .\"" + }, + { + "Answer": 0.16666666666666666, + "Options": "a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 9 , e ) 1 / 12", + "Correct": "c", + "Explanation": "there are 6 ways to roll a seven : 1 and 6 , 6 and 1 , 2 and 5 , 5 and 2 , 3 and 4 , 4 and 3 . there are 6 * 6 = 36 ways to roll two six - sided dice . thus the probability of winning by rolling a seven on the second roll given a seven on the first roll is 6 / 36 = 1 / 6 c" + }, + { + "Answer": 0.4166666666666667, + "Options": "a ) 3 / 10 , b ) 4 / 11 , c ) 5 / 12 , d ) 6 / 13 , e ) 7 / 15", + "Correct": "c", + "Explanation": "the total possible ways of selecting a 6 - member team is 9 c 6 = 84 the possible ways which include john and peter is 7 c 4 = 35 the probability of choosing both john and peter is 35 / 84 = 5 / 12 the answer is c ." + }, + { + "Answer": 144, + "Options": "a ) 144 , b ) 288 , c ) 12 , d ) 256 , e ) none", + "Correct": "a", + "Explanation": "\"solution : let the arrangement be , b g b g b g b 4 boys can be seated in 4 ! ways . girl can be seated in 3 ! ways . required number of ways , = 4 ! * 3 ! = 144 . answer : option a\"" + }, + { + "Answer": 0.125, + "Options": "a ) 1 / 8 , b ) 1 / 6 , c ) 1 / 4 , d ) 3 / 4 , e ) 7 / 8", + "Correct": "a", + "Explanation": "good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 3 / 4 of the garden is green , so , since all the flowers must be either green or yellow , we know that 1 / 4 are yellow . we ' re also told there is an equal probability of straight or curved , 1 / 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( curved ) = 1 / 4 * 1 / 2 = 1 / 8 , or a ." + }, + { + "Answer": 6, + "Options": "a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8", + "Correct": "c", + "Explanation": "explanation : clearly , the black cards are either clubs or spades while the red cards are either diamonds or hearts . let the number of spades be x . then , number of clubs = ( 7 - x ) . number of diamonds = 2 x number of spades = 2 x ; number of hearts = 2 x number of diamonds = 4 x . total number of cards = x + 2 x + 4 x + 7 - x = 6 x + 7 . therefore 6 x + 7 = 13 6 x = 6 x - 1 . hence , number of clubs = ( 7 - x ) = 6 . answer : c" + }, + { + "Answer": 0.9, + "Options": "a ) 3 / 4 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 9 / 10", + "Correct": "e", + "Explanation": "\"bag consists of 50 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 50 - 5 - 9 = 36 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 50 + 36 / 50 = 45 / 50 = 9 / 10 hence , answer will be e .\"" + }, + { + "Answer": 0.3241106719367589, + "Options": "a ) 4 / 19 , b ) 7 / 19 , c ) 12 / 37 , d ) 21 / 95 , e ) none", + "Correct": "c", + "Explanation": "solution p ( none is defective ) = 19 c 2 / 23 c 2 = 25 / 37 . p ( at least one is defective ) = ( 1 - 25 / 37 ) = 12 / 37 . answer c" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 10 / 30 , b ) 20 / 30 , c ) 1 , d ) 5 / 30 , e ) 15 / 30", + "Correct": "a", + "Explanation": "x = { 2,3 , 5,7 , 11,13 , 17,19 , 23,29 } n ( x ) = 10 n ( s ) = 30 hence required probability = n ( x ) / n ( s ) = 10 / 30 option a is answer" + }, + { + "Answer": 0.5142857142857142, + "Options": "a ) 18 / 35 , b ) 16 / 35 , c ) 14 / 35 , d ) 13 / 35 , e ) 12 / 35", + "Correct": "a", + "Explanation": "men = women means 2 men 2 women total such combinations = 4 c 2 * 4 c 2 = 4 ! / 2 ! . 2 ! * 4 ! / 2 ! . 2 ! = 6 * 6 total combinations = 8 c 4 = 8 * 7 * 6 * 5 / 4 * 3 * 2 * 1 = 70 so probability = 36 / 70 = 18 / 35 hence a" + }, + { + "Answer": 0.9666666666666667, + "Options": "a ) 1 / 6 , b ) 4 / 120 , c ) 29 / 30 , d ) 1 / 9 , e ) none of these", + "Correct": "c", + "Explanation": "explanation : out of 10 electric bulbs , 3 are selected in 10 c 3 ways . so , number of possible outcomes = t = 10 c 3 = 120 . the room is lighted i . e at least one fo the 3 bulbs chosen at random is a good bulb . wherever ' at least one ' is present , it is easier to do the problem using complement of the event ( i . e room is dark and all 3 bulbs are bad ) . p ( room is dark ) = f / t = p ( none of the bulbs is good ) = > 4 c 3 / 120 . [ since 4 bulbs are bad ] . p ( room is lighted ) = 1 - ( 4 c 3 / 120 ) . [ room is lighted if at least one bulb is good ] . or p ( at least one ) = 1 - ( 4 / 120 ) . = 29 / 30 . answer : c" + }, + { + "Answer": 0.625, + "Options": "a ) 2 / 3 , b ) 5 / 8 , c ) 5 / 9 , d ) 3 / 7 , e ) 4 / 7", + "Correct": "b", + "Explanation": "\"since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . \u2234 required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b\"" + }, + { + "Answer": 0.15625, + "Options": "a ) 5 / 32 , b ) 6 / 32 , c ) 5 / 16 , d ) 4 / 32 , e ) 8 / 32", + "Correct": "a", + "Explanation": "exactly 4 heads out of 5 would be : hhhht , hhthh , or many other combinations . therefore , no . of possible combinations are 5 ! / 4 ! * 1 ! = 5 probability = 5 / 2 ^ 5 = 5 / 32 answer : a" + }, + { + "Answer": 0.6923076923076923, + "Options": "a ) 9 / 13 , b ) 5 / 12 , c ) 3 / 13 , d ) 6 / 17 , e ) 1 / 13", + "Correct": "a", + "Explanation": "clearly in the 52 cards other than face cards = 52 - 16 = 36 probability of not getting a face card = 36 / 52 = 9 / 13 correct option is a" + }, + { + "Answer": 0.11904761904761904, + "Options": "a ) 2 / 3 , b ) 1 / 3 , c ) 1 / 5 , d ) 5 / 42 , e ) 7 / 42", + "Correct": "d", + "Explanation": "combination probability formula : ncr = n ! / [ r ! ( n - r ) ! ] total possible , select 3 bags from 9 bags = 9 c 3 = 9 ! / [ 3 ! ( 9 - 3 ) ! ] = 84 . to have 3 bbq there must be 0 plain , select 0 plain from 4 = 4 c 0 = 1 . and , select 3 bbq from 5 = 5 c 3 = 10 . 3 bbq and 0 plain = ( 1 ) ( 10 ) = 10 probability = ( number outcomes favorable ) / ( total number outcomes ) = 10 / 84 = 5 / 42 answer : d" + }, + { + "Answer": 0.5555555555555556, + "Options": "a ) 5 / 8 , b ) 5 / 6 , c ) 5 / 1 , d ) 5 / 9 , e ) 5 / 2", + "Correct": "d", + "Explanation": "bag a 6 red + 3 pink bag b 2 red + 7 pink probability of picking 1 pink from bag a is 3 / 9 probability of picking 1 pink from bag b is 7 / 9 total = 3 / 9 + 7 / 9 = 10 / 9 and probability of selecting a bag is 1 / 2 so finally probability of choosing a pink flower = 1 / 2 ( 10 / 9 ) = 5 / 9 answer : d" + }, + { + "Answer": 0.2857142857142857, + "Options": "a ) 2 / 7 , b ) 2 / 6 , c ) 2 / 5 , d ) 2 / 3 , e ) 2 / 4", + "Correct": "a", + "Explanation": "\"a leap year has 366 day which is 52 full weeks + 2 odd days . now these two odd days may be ( sun + mon ) , ( mon + tue ) , . . . . ( sat + sun ) . now there are total 7 ways . of which sunday appeared two times . so answer 2 / 7 answer : a\"" + }, + { + "Answer": 0.4, + "Options": "a ) 5 / 5 , b ) 4 / 5 , c ) 3 / 5 , d ) 2 / 5 , e ) 1 / 5", + "Correct": "d", + "Explanation": "this can be solved in a generic and complex way but let us not go into all that . there can be four ways through which the pair of dice results in a sum of 5 . there can be six ways through which the pair of dice can result in a sum of 7 . now , we want the probability of the pair of dice resulting in a sum of 5 before a sum of 7 . thus probability = 4 / ( 4 + 6 ) = 4 / 10 or 2 / 5 . correct answer is d ) 2 / 5" + }, + { + "Answer": 0.234375, + "Options": "a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64", + "Correct": "a", + "Explanation": "\"chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days q = 15 / 64 ans a it is .\"" + }, + { + "Answer": 0.234375, + "Options": "a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64", + "Correct": "a", + "Explanation": "\"chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days e = 15 / 64 ans a it is .\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 2 , d ) 1 / 5 , e ) 1 / 6", + "Correct": "a", + "Explanation": "\"taking 2 red shoe the probablity is 6 c 2 from 10 shoes probablity of taking 2 red shoe is 6 c 2 / 10 c 2 = 1 / 3 answer : a\"" + }, + { + "Answer": 0.48, + "Options": "a ) 8 / 25 , b ) 11 / 24 , c ) 12 / 25 , d ) 12 / 25 , e ) 12 / 23", + "Correct": "d", + "Explanation": "explanation : p ( carol ) = 4 / 5 p ( bernie ) = 3 / 5 e = { carol and bernie both get selected } p ( e ) = p ( carol ) * p ( bernie ) = 4 / 5 * 3 / 5 = 12 / 25 answer : d" + }, + { + "Answer": 0.3076923076923077, + "Options": "a ) 4 / 15 , b ) 4 / 10 , c ) 4 / 18 , d ) 4 / 13 , e ) 4 / 11", + "Correct": "d", + "Explanation": "\"p ( s \u1d1c k ) = p ( s ) + p ( k ) - p ( s \u2229 k ) , where s denotes spade and k denotes king . p ( s \u1d1c k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : d\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 9", + "Correct": "b", + "Explanation": "\"from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 6 = 1 / 3 . the answer is b .\"" + }, + { + "Answer": 0.5357142857142857, + "Options": "a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14", + "Correct": "d", + "Explanation": "\"the required probability z = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer .\"" + }, + { + "Answer": 0.0625, + "Options": "a ) 1 / 5 , b ) 1 / 10 , c ) 1 / 12 , d ) 1 / 16 , e ) 1 / 32", + "Correct": "d", + "Explanation": "\"must be either 5 heads or 5 tails 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 + 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 1 / 32 + 1 / 32 = 1 / 16 answer : d\"" + }, + { + "Answer": 90, + "Options": "a ) 90 , b ) 85 , c ) 75 , d ) 95 , e ) 82", + "Correct": "a", + "Explanation": "first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 10 and for second = 9 so total methods = 10 * 9 = 90 answer : a" + }, + { + "Answer": 1540, + "Options": "a ) 1540 , b ) 2314 , c ) 5420 , d ) 1240 , e ) 4512", + "Correct": "a", + "Explanation": "\"we must place all these books as under : xexex \u2026 . xex there are 21 books on english , 19 books on hindi 19 places out of 22 can be chosen 22 c 19 = 22 c 3 = 1540 ways ans : a\"" + }, + { + "Answer": 0.007936507936507936, + "Options": "a ) 1 / 126 , b ) 1 / 120 , c ) 1 / 24 , d ) 4 / 25 , e ) 1 / 2", + "Correct": "a", + "Explanation": "\"consider a case when all girls select black marbles . . . so girl 1 will chose 1 black marble out of 10 . . and black marble will be chosen from 5 black marbles girl 1 will have a probability of picking black marble 5 / 10 girl 2 will have to pick a black marble out of 4 and total remaining no . of marbles 9 = 4 / 9 so we have probability of all girls selecting black marbles as 5 / 10 * 4 / 9 * 3 / 8 * 2 / 7 * 1 / 6 = 1 / 252 since girls can also select white marbles so we will have 2 cases so probability of girls selecting all same colour marbles is 2 * 1 / 252 or 1 / 126 ans is a\"" + }, + { + "Answer": 0.5, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 7", + "Correct": "a", + "Explanation": "\"from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 4 = 1 / 2 . the answer is a .\"" + }, + { + "Answer": 0.26666666666666666, + "Options": "a ) 2 / 15 , b ) 4 / 15 , c ) 1 / 15 , d ) 4 / 90 , e ) 5 / 60", + "Correct": "b", + "Explanation": "solution : there are a total of 90 two digit numbers . every third number will be divisible by ' 3 ' . therefore , there are 30 of those numbers that are divisible by ' 3 ' . of these 30 numbers , the numbers that are divisible by ' 5 ' are those that are multiples of ' 15 ' . i . e . numbers that are divisible by both ' 3 ' and ' 5 ' . there are 6 such numbers - - 15 , 30,45 , 60,75 and 90 . we need to find out numbers that are divisible by ' 3 ' and not by ' 5 ' , which will be : 30 \u00e2 \u02c6 \u2019 6 = 24 24 out of the 90 numbers are divisible by ' 3 ' and not by ' 5 ' . the required probability is therefore , = 24 / 90 = 4 / 15 answer is b" + }, + { + "Answer": 0.25, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 11", + "Correct": "c", + "Explanation": "\"from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 8 = 1 / 4 . the answer is c .\"" + }, + { + "Answer": 0.5, + "Options": "a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 2 / 5 , e ) 1 / 2", + "Correct": "e", + "Explanation": "\"s = { h , t } e = { h } probability = n ( e ) / n ( s ) = 1 / 2 correct option is e\"" + }, + { + "Answer": 0.2857142857142857, + "Options": "a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 2 / 7", + "Correct": "e", + "Explanation": "\"there are 366 days in a leap year : 52 weeks and 2 more days . so , 52 sundays and 2 days . these 2 days can be : { mon , tue } , { tue , wed } , { wed , thu } , { thu , fri } , { fri , sat } , { sat , sun } and { sun , mon } ( 7 cases ) . in order to have 53 sundays we should have either { sat , sun } or { sun , mon } case . the probability of that is 2 / 7 . answer : e .\"" + }, + { + "Answer": 0.13333333333333333, + "Options": "a ) 1 / 225 , b ) 1 / 30 , c ) 1 / 15 , d ) 2 / 15 , e ) 4 / 15", + "Correct": "d", + "Explanation": "g 1 / 15 + ( 14 / 15 ) * 1 / 14 = 2 / 15 d" + }, + { + "Answer": 25.00000000000001, + "Options": "a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 %", + "Correct": "b", + "Explanation": "\"could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . y = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %\"" + }, + { + "Answer": 0.027777777777777776, + "Options": "a ) 1 / 144 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 3", + "Correct": "b", + "Explanation": "\"in order to get the sum of the two dices to be 7 , there are six combinations that satisfy this requirement . 1 + 6 = 7 2 + 5 = 7 3 + 4 = 7 4 + 3 = 7 5 + 2 = 7 6 + 1 = 7 each combination has 1 / 36 of probability of occurrence , or in total 6 / 36 for the six combinations , which is 1 / 6 . in order to get the probability of this occurring twice in a row we multiply 1 / 6 * 1 / 6 and we get 1 / 36 . the answer is b .\"" + }, + { + "Answer": 0.3666666666666667, + "Options": "a ) 1 / 3 , b ) 2 / 3 , c ) 11 / 120 , d ) 11 / 30 , e ) 76 / 120", + "Correct": "d", + "Explanation": "this q is based on derangement , a permutation in which all elements are in the wrong position . number of derangements = n ! ( 12 ! \u2212 13 ! + 14 ! + . . . + ( ( \u2212 1 ) n ) n ! ) n ! ( 12 ! \u2212 13 ! + 14 ! + . . . + ( ( \u2212 1 ) n ) n ! ) . . of course it can be derived but thats not required . . since there are 5 letters and 5 envelopes : - derangements = 5 ! ( 12 ! \u2212 13 ! + 14 ! \u2212 15 ! ) derangements = 5 ! ( 12 ! \u2212 13 ! + 14 ! \u2212 15 ! ) = > 120 ( 1 / 2 \u2212 1 / 6 + 1 / 24 \u2212 1 / 120 ) = 60 \u2212 20 + 5 \u2212 1 = 44.120 ( 1 / 2 \u2212 1 / 6 + 1 / 24 \u2212 1 / 120 ) = 60 \u2212 20 + 5 \u2212 1 = 44 . total possible ways = 5 ! = 120 . therefore required prob , p = 44 / 120 = 11 / 30 . ans d" + }, + { + "Answer": 0.16666666666666669, + "Options": "a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 4 , d ) 3 / 4 , e ) 7 / 8", + "Correct": "a", + "Explanation": "\"good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 2 / 3 of the garden is green , so , since all the flowers must be either green or yellow , we know that 1 / 3 are yellow . we ' re also told there is an equal probability of straight or curved , 1 / 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( straight ) = 1 / 3 * 1 / 2 = 1 / 6 , or a .\"" + }, + { + "Answer": 0.5, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 2 / 3 , e ) 2 / 5", + "Correct": "a", + "Explanation": "s = { h , t } e = { h } p ( e ) = 1 / 2 answer is a" + }, + { + "Answer": 25.00000000000001, + "Options": "a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 %", + "Correct": "b", + "Explanation": "could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . r = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %" + }, + { + "Answer": 0.0625, + "Options": "a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 2 , d ) 1 / 16 , e ) 1 / 32", + "Correct": "d", + "Explanation": "( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 16 answer : d" + }, + { + "Answer": 0.09090909090909091, + "Options": "a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 11 , e ) 1 / 25", + "Correct": "d", + "Explanation": "total pairs = 12 c 2 = 66 ; same color pairs = 6 c 1 * 1 c 1 = 6 ; prob = 1 / 11 ans d" + }, + { + "Answer": 0.6666666666666666, + "Options": "a ) 2 / 3 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these", + "Correct": "a", + "Explanation": "\"explanation : number greater than 2 is 3 , 4 , 5 & 6 , so only 4 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 4 / 6 = 2 / 3 answer : a\"" + }, + { + "Answer": 0.14285714285714285, + "Options": "a ) 1 / 7 , b ) 1 / 8 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 25", + "Correct": "a", + "Explanation": "\"total pairs = 8 c 2 = 28 ; same color pairs = 4 c 1 * 1 c 1 = 4 ; prob = 1 / 7 ans a\"" + }, + { + "Answer": 6600, + "Options": "a ) 6600 , b ) 1300 , c ) 6780 , d ) 1976 , e ) 2448", + "Correct": "a", + "Explanation": "\"e = event that 3 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 11 and 3 girl from 10 = 11 c 2 * 10 c 3 = 6600 ans - a\"" + }, + { + "Answer": 780, + "Options": "a ) 120 , b ) 480 , c ) 780 , d ) 720 , e ) 800", + "Correct": "c", + "Explanation": "n ( s ) = sample space = 23 c 3 = 1771 e = event that 1 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 13 and 1 girl from 10 = 13 c 2 * 10 c 1 = 780 ans - c" + }, + { + "Answer": 5, + "Options": "a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5", + "Correct": "e", + "Explanation": "given s = 3 t t can take 0,1 , 2,3 , 4 and so s = 0,3 , 6,9 , 12 4 such pairs can be formed . answer e" + }, + { + "Answer": 0.875, + "Options": "a ) 7 / 8 , b ) 5 / 8 , c ) 3 / 8 , d ) 7 / 5 , e ) 3 / 5", + "Correct": "a", + "Explanation": "\"here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting at least two heads = { ttt , tth , tht , htt , thh , hth , hht } p ( e ) = n ( e ) / n ( s ) = 7 / 8 . answer a .\"" + }, + { + "Answer": 0.7, + "Options": "a ) 7 / 10 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16", + "Correct": "a", + "Explanation": "\"bag consists of 20 marbles , of which 6 are blue , 9 are red remainder are white . so , white = 20 - 6 - 9 = 5 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 20 + 5 / 20 = 14 / 20 = 7 / 10 hence , answer will be a .\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 2 , d ) 3 / 4 , e ) 3 / 7", + "Correct": "a", + "Explanation": "explanation : let 1 st girl come and she choose wrong dress so probability of that girl to choose wrong dress out of 3 is = 2 / 3 . now 2 nd girl come nd she choose wrong dress so probability of that girl to choose wrong dress out of 2 is 1 / 2 . now for 3 rd girl probability is 1 to choose wrong dress . so probability tht all the 3 wil not choose der own dress is = 2 / 3 * 1 / 2 * 1 = 1 / 3 . hence ( a ) is the correct answer . answer : a" + }, + { + "Answer": 3003, + "Options": "a ) 210 , b ) 1365 , c ) 3003 , d ) 5005 , e ) 3030", + "Correct": "c", + "Explanation": "the number of ways of choosing 5 toys out of 15 are counted by = 15 c 5 ways = 3003 answer : option c" + }, + { + "Answer": 0.21818181818181817, + "Options": "a ) 9 / 16 , b ) 7 / 18 , c ) 12 / 55 , d ) 14 / 67 , e ) 19 / 87", + "Correct": "c", + "Explanation": "total number of balls = 3 + 5 + 4 = 12 n ( s ) = 12 c 3 = 12 * 11 * 10 / 3 * 2 = 220 n ( e ) = 4 c 2 * 8 c 1 = 6 * 8 = 48 probability = 48 / 220 = 12 / 55 answer is c" + }, + { + "Answer": 0.16666666666666666, + "Options": "a ) 1 / 18 , b ) 1 / 12 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 6", + "Correct": "e", + "Explanation": "explanation : there are total 3 letters . the probability that o gets the first position is 1 / 3 . likewise , probability for l and t are 1 / 2 and 1 / 1 . hence , the required probability is : - = > ( 1 / 3 ) x ( 1 / 2 ) x 1 . = > 1 / 6 . answer : e" + }, + { + "Answer": 145.8, + "Options": "a ) 140 , b ) 146 , c ) 152 , d ) 162 , e ) 170", + "Correct": "b", + "Explanation": "\"number of newborns that can die in first month = 1 / 10 * 200 = 20 survived = 180 number of newborns that can die in second month = 1 / 10 * 180 = 18 survived = 162 number of newborns that can die in third month = 1 / 10 * 162 = 16 survived = 146 answer : b\"" + }, + { + "Answer": 0.23809523809523808, + "Options": "a ) 2 / 7 , b ) 5 / 21 , c ) 4 / 21 , d ) 3 / 7 , e ) 1 / 2", + "Correct": "b", + "Explanation": "p ( 1 st black , 2 nd white ) = 5 / 7 * 2 / 6 = 10 / 42 ; p ( 1 st white , 2 nd black ) = 2 / 7 * 5 / 6 = 10 / 42 . p = 10 / 42 + 10 / 42 = 20 / 84 = 10 / 42 = 5 / 21 answer : b ." + }, + { + "Answer": 6, + "Options": "a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6", + "Correct": "e", + "Explanation": "x / 15 = 1 / 5 x = 3 3 + 3 ( removed 3 blue balls ) = 6 answer : e" + }, + { + "Answer": 0.625, + "Options": "a ) 2 / 3 , b ) 5 / 8 , c ) 5 / 4 , d ) 5 / 6 , e ) 6 / 7", + "Correct": "b", + "Explanation": "\"since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . \u2234 required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 3 , b ) 1 / 4 , c ) 2 / 3 , d ) 2 / 5 , e ) 3 / 7", + "Correct": "a", + "Explanation": "s = { 1,2 , 3,4 , 5,6 } e = { 5,6 } probability = 2 / 6 = 1 / 3 answer is a" + }, + { + "Answer": 0.03125, + "Options": "a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 16 , d ) 1 / 32 , e ) 1 / 64", + "Correct": "d", + "Explanation": "\"p ( htttt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 the answer is d .\"" + }, + { + "Answer": 0.25, + "Options": "a ) 1 / 4 , b ) 3 / 40 , c ) 13 / 40 , d ) 7 / 20 , e ) 13 / 22", + "Correct": "a", + "Explanation": "p ( 1 and not 2 ) = 5 / 8 * ( 1 - 3 / 5 ) = 1 / 4 answer : a ." + }, + { + "Answer": 0.03125, + "Options": "a ) 3 / 5 , b ) 1 / 32 , c ) 1 / 5 , d ) 1 / 4 , e ) 1 / 2", + "Correct": "b", + "Explanation": "( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 32 answer : b" + }, + { + "Answer": 0.8461538461538461, + "Options": "a ) 12 / 13 , b ) 11 / 13 , c ) 10 / 15 , d ) 1 / 15 , e ) 4 / 14", + "Correct": "b", + "Explanation": "total possible ways of selecting 4 students out of 15 = 15 c 4 = ( 15 * 14 * 13 * 12 ) / ( 4 * 3 * 2 * 1 ) = 1365 the number of ways of selecting a 4 students in which no student belongs to karnataka = 10 c 4 number of ways of selecting atleast one student from karnataka = 15 c 4 - 10 c 4 = 1155 probability = 1155 / 1365 = 11 / 13 answer is b" + }, + { + "Answer": 120, + "Options": "a ) 20 , b ) 120 , c ) 210 , d ) 15 , e ) 45", + "Correct": "b", + "Explanation": "\"the number of ways of choosing 3 toys out of 10 are counted by = 10 c 3 ways = 120 answer : option b\"" + }, + { + "Answer": 0.8333333333333334, + "Options": "a ) 1 / 3 , b ) 4 / 9 , c ) 5 / 6 , d ) 3 / 7 , e ) 1 / 4", + "Correct": "c", + "Explanation": "\"total possible ways of selecting 4 students out of 10 = 10 c 4 = ( 10 * 9 * 8 * 7 ) / ( 4 * 3 * 2 * 1 ) = 210 the number of ways of selecting a 4 students in which no student belongs to karnataka = 7 c 4 number of ways of selecting atleast one student from karnataka = 15 c 4 - 7 c 4 = 175 probability = 175 / 210 = 5 / 6 answer is c\"" + }, + { + "Answer": 1050, + "Options": "a ) 950 , b ) 1050 , c ) 2150 , d ) 2050 , e ) 1000", + "Correct": "b", + "Explanation": "\"n ( s ) = sample space = 25 c 3 = 2300 e = event that 1 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 15 and 1 girl from 10 = 15 c 2 * 10 c 1 = 1050 ans - b\"" + }, + { + "Answer": 17280, + "Options": "a ) 14420 , b ) 12620 , c ) 17280 , d ) 21200 , e ) 24560", + "Correct": "c", + "Explanation": "\"6 boys can be seated in 6 ! 4 girls can be seated in 4 ! required number = 6 ! 4 ! = 17280 answer is c\"" + }, + { + "Answer": 1771, + "Options": "a ) 4586 , b ) 5896 , c ) 2415 , d ) 1771 , e ) 1881", + "Correct": "d", + "Explanation": "\"in order that two books on hindi are never together , we must place all these books as under : h e h e h e h \u2026 . . h e h where h denotes the position of hindi book and e that of english book . since there are 22 books on english , the number of places marked e are 23 . now , 20 places out of 23 can be chosen in 23 c 20 = 23 c 3 = 23 * 22 * 21 3 * 2 * 1 = 1771 ways . hence the number of ways = 1771 ways answer : d\"" + }, + { + "Answer": 0.6666666666666666, + "Options": "a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1", + "Correct": "d", + "Explanation": "\"answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability j is 2 / 3 . d\"" + }, + { + "Answer": 0.1, + "Options": "a ) 1 / 35 , b ) 1 / 10 , c ) 1 / 15 , d ) 1 / 20 , e ) 1 / 25", + "Correct": "b", + "Explanation": "first , total ways to select for all boys and girls , i . e 6 ! / ( 3 ! * 3 ! ) = 6 * 5 * 4 * 3 * 2 * 1 / 3 * 2 * 1 * 3 * 2 * 1 = 20 then there are one two way girls can have all same colors , either white or black . the number of ways in which 3 girls can select 3 white balls = 3 c 3 = 1 the number of ways in which 3 girls can select 3 black balls = 3 c 3 = 1 therefore , total favorable outcomes / total outcomes = 2 / 20 = 1 / 10 b" + }, + { + "Answer": 437.40000000000003, + "Options": "a ) 438 , b ) 446 , c ) 452 , d ) 462 , e ) 470", + "Correct": "a", + "Explanation": "\"number of newborns that can die in first month = 1 / 10 * 600 = 60 survived = 540 number of newborns that can die in second month = 1 / 10 * 540 = 54 survived = 486 number of newborns that can die in third month = 1 / 10 * 486 = 48 survived = 438 answer : a\"" + }, + { + "Answer": 0.0007716049382716049, + "Options": "a ) 1 / 32 , b ) 1 / 1296 , c ) 1 / 33 , d ) 1 / 38 , e ) 1 / 34", + "Correct": "b", + "Explanation": "the three dice can fall in 6 * 6 * 6 * 6 * 6 = 7776 ways . hence the probability is 6 / 7776 = 1 / 1296 answer : b" + }, + { + "Answer": 6, + "Options": "a ) 5 , b ) 1 , c ) 3 , d ) 6 , e ) 7", + "Correct": "d", + "Explanation": "\"first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 3 and for second = 2 so total methods = 3 * 2 = 6 answer : d\"" + }, + { + "Answer": 0.6666666666666666, + "Options": "a ) 2 / 3 , b ) 1 / 2 , c ) 3 / 4 , d ) 5 / 6 , e ) 1 / 3", + "Correct": "a", + "Explanation": "the numbers which satisfy the conditions are 1 , 2 , 4 , and 6 . the probability is 4 / 6 = 2 / 3 the answer is a ." + }, + { + "Answer": 0.368421052631579, + "Options": "a ) 8 / 19 , b ) 7 / 19 , c ) 10 / 19 , d ) 20 / 19 , e ) 22 / 19", + "Correct": "b", + "Explanation": "\"p ( none is defective ) = 16 c 2 / 20 c 2 = 12 / 19 p ( at least one is defective ) = 1 - 12 / 19 = 7 / 19 answer b\"" + }, + { + "Answer": 90, + "Options": "a ) 30 , b ) 60 , c ) 90 , d ) 180 , e ) 540", + "Correct": "c", + "Explanation": "90 is the number of ways you can assign 3 teams formed out of 12 people to 3 different tasks . but now you can order the 3 tasks in 3 ! ways . t 1 t 2 t 3 or t 2 t 1 t 3 . . . . etc etc . i was confused between 90 and 540 but since question used the wordarrangementsdecided to go with complete arrangements w including the order of tasks . could you explain the highlighted step . . . i ' m getting 90 = 15 * 3 ! suppose the students are numbered 12 , 34 , 56 and tasks are x , y and z one of the 15 possible ways of forming teams is 12 , 34 , 56 . these teams can be assigned to 3 tasks in 3 ! = 6 ways x - - y - - z 12 - - 34 - - 56 12 - - 56 - - 34 34 - - 12 - - 56 34 - - 56 - - 12 56 - - 12 - - 34 56 - - 34 - - 12 so the answer should be 15 * 6 = 90 but now you can fruther decide which task you want to perform first x y or z . . = c" + }, + { + "Answer": 42, + "Options": "a ) 50 , b ) 42 , c ) 24 , d ) 20 , e ) 12", + "Correct": "b", + "Explanation": "\"21 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 21 x 2 = 42 b\"" + }, + { + "Answer": 0.3, + "Options": "a ) 17 / 30 , b ) 2 / 5 , c ) 3 / 10 , d ) 4 / 15 , e ) 11 / 30", + "Correct": "c", + "Explanation": "number of multiples of 4 from 1 through 30 = 30 / 4 = 7 number of multiples of 15 from 1 through 30 = 30 / 15 = 2 number of multiples of 4 and 15 both from 1 through 30 = number of multiples of 15 * 4 ( = 60 ) = 0 total favourable cases = 7 + 2 - 0 = 9 probability = 9 / 30 = 3 / 10 answer : option c" + }, + { + "Answer": 0.625, + "Options": "a ) 2 / 3 , b ) 5 / 8 , c ) 3 / 5 , d ) 3 / 8 , e ) 4 / 9", + "Correct": "b", + "Explanation": "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . \u2234 required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b" + }, + { + "Answer": 0.15000000000000002, + "Options": "a ) 1 / 20 , b ) 3 / 40 , c ) 13 / 40 , d ) 3 / 20 , e ) 13 / 22", + "Correct": "d", + "Explanation": "\"p ( 1 and not 2 ) = 3 / 8 * ( 1 - 3 / 5 ) = 14 / 40 = 3 / 20 . answer : d .\"" + }, + { + "Answer": 0.3115942028985508, + "Options": "a ) 4 / 19 , b ) 7 / 19 , c ) 12 / 19 , d ) 24 / 77 , e ) none", + "Correct": "d", + "Explanation": "\"solution p ( none is defective ) = 20 c 2 / 24 c 2 = 53 / 77 . p ( at least one is defective ) = ( 1 - 53 / 77 ) = 24 / 77 . answer d\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 2 / 4 , b ) 1 / 3 , c ) 2 / 3 , d ) 3 / 6 , e ) 1 / 4", + "Correct": "b", + "Explanation": "sample space s = { hh , ht , t 1 , t 2 , t 3 , t 4 , t 5 , t 6 } let a be the event that the die shows a number greater than 4 and b be the event that the first throw of the coin results in a tail then , a = { t 5 , t 6 } b = { t 1 , t 2 , t 3 , t 4 , t 5 , t 6 } required probability = ( 2 / 8 ) / ( 6 / 8 ) = 1 / 3 answer : b" + }, + { + "Answer": 0.47619047619047616, + "Options": "a ) 13 / 25 , b ) 2 / 5 , c ) 9 / 20 , d ) 11 / 20 , e ) 10 / 21", + "Correct": "e", + "Explanation": "explanation : p ( jamie ) = 2 / 3 p ( tom ) = 5 / 7 e = { jamie and tom both get selected } p ( e ) = p ( jamie ) * p ( tom ) = 2 / 3 * 5 / 7 = 10 / 21 answer : e" + }, + { + "Answer": 0.09375, + "Options": "a ) 3 / 8 , b ) 3 / 16 , c ) 5 / 16 , d ) 3 / 32 , e ) 7 / 64", + "Correct": "d", + "Explanation": "\"the number of possible outcomes is 2 ^ 6 = 64 there are 6 ways to get exactly 5 heads . p ( exactly 5 heads ) = 6 / 64 = 3 / 32 the answer is d .\"" + }, + { + "Answer": 291.6, + "Options": "a ) 240 , b ) 246 , c ) 252 , d ) 292 , e ) 298", + "Correct": "d", + "Explanation": "\"number of newborns that can die in first month = 1 / 10 * 400 = 40 survived = 360 number of newborns that can die in second month = 1 / 10 * 360 = 36 survived = 324 number of newborns that can die in third month = 1 / 10 * 324 = 32 survived = 292 answer : d\"" + }, + { + "Answer": 0.45454545454545453, + "Options": "a ) 3 / 4 , b ) 5 / 11 , c ) 1 / 7 , d ) 1 / 8 , e ) 4 / 3", + "Correct": "b", + "Explanation": "\"let number of balls = ( 6 + 5 ) = 11 . number of white balls = 5 p ( drawing a white ball ) = 5 / 11 option b .\"" + }, + { + "Answer": 48, + "Options": "a ) 48 , b ) 288 , c ) 12 , d ) 256 , e ) none", + "Correct": "a", + "Explanation": "\"solution : let the arrangement be , b g b g b b 4 boys can be seated in 4 ! ways . girl can be seated in 2 ! ways . required number of ways , = 4 ! * 2 ! = 48 answer : option a\"" + }, + { + "Answer": 0.10526315789473684, + "Options": "a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10", + "Correct": "b", + "Explanation": "\"method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability t = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b\"" + }, + { + "Answer": 25, + "Options": "a ) 25 , b ) 20 , c ) 35 , d ) 36 , e ) 38", + "Correct": "a", + "Explanation": "\"we can either choose all 4 people from 5 manager who have no problems or choose 3 from the 5 and 1 from the 2 managers who have a problem sitting together so 5 c 4 + ( 5 c 3 * 2 c 1 ) this is 5 + 20 = 25 answer : a\"" + }, + { + "Answer": 14, + "Options": "a ) a ) 40 , b ) b ) 14 , c ) c ) 50 , d ) d ) 55 , e ) e ) 60", + "Correct": "b", + "Explanation": "\"approach 1 at - least 1 paper back = total - no paper back 6 c 4 - 4 c 4 = 14 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 2 c 1 * 4 c 3 + 2 c 2 * 4 c 2 = 14 answer is b\"" + }, + { + "Answer": 0.21428571428571427, + "Options": "a ) 2 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 15 , e ) 3 / 14", + "Correct": "e", + "Explanation": "\"p ( 1 st black , 2 nd white ) = 6 / 8 * 2 / 7 = 12 / 56 ; p ( 1 st white , 2 nd black ) = 2 / 8 * 6 / 7 = 12 / 56 . p = 12 / 56 + 12 / 56 = 24 / 112 = 12 / 56 = 6 / 28 = 3 / 14 . answer : e .\"" + }, + { + "Answer": 65, + "Options": "a ) 22 , b ) 87 , c ) 29 , d ) 26 , e ) 191", + "Correct": "b", + "Explanation": "answer : b" + }, + { + "Answer": 2, + "Options": "a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 12", + "Correct": "a", + "Explanation": "\"1 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 1 x 2 = 2 a\"" + }, + { + "Answer": 0.6666666666666666, + "Options": "a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these", + "Correct": "b", + "Explanation": "explanation : number greater than 2 is 3 , 4 , 5 & 6 , so only 4 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 4 / 6 = 2 / 3 answer : b" + }, + { + "Answer": 0.68, + "Options": "a ) 21 / 25 , b ) 17 / 25 , c ) 4 / 25 , d ) 8 / 25 , e ) 10 / 25", + "Correct": "b", + "Explanation": "\"the number of exhaustive events = 50 c 1 = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . therefore , required probability = 34 / 50 = 17 / 25 . answer : b\"" + }, + { + "Answer": 0.2777777777777778, + "Options": "a ) 1 / 18 , b ) 1 / 14 , c ) 5 / 18 , d ) 1 / 15 , e ) 1 / 16", + "Correct": "c", + "Explanation": "taking 2 red shoe the probability is 5 c 2 from 9 shoes probability of taking 2 red shoes is 5 c 2 / 9 c 2 = 5 / 18 answer : c" + }, + { + "Answer": 0.42857142857142855, + "Options": "a ) 1 , b ) 2 , c ) none , d ) 3 / 7 , e ) 3 / 4", + "Correct": "d", + "Explanation": "the total number of candies is 7 and the number of candies with lemon flavor is 3 then , in one chance the probability to pick one candie with lemon flavor is 3 / 7 answer d" + }, + { + "Answer": 0.375, + "Options": "a ) 0.125 , b ) 0.25 , c ) 0.75 , d ) 0.375 , e ) 0.666", + "Correct": "d", + "Explanation": "\"total number of ways in which h or t can appear in 3 tosses of coin is = 2 * 2 * 2 = 8 ways for 2 t and 1 h thus probability is = p ( htt ) + p ( tth ) + p ( tht ) = 1 / 8 + 1 / 8 + 1 / 8 = 3 / 8 = . 375 answer : d\"" + }, + { + "Answer": 0.375, + "Options": "a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 8 , d ) 5 / 8 , e ) 3 / 16", + "Correct": "c", + "Explanation": "\"the number of possible outcomes is 2 ^ 3 = 8 there are 3 ways to get exactly 2 heads . p ( exactly 2 heads ) = 3 / 8 the answer is c .\"" + }, + { + "Answer": 36, + "Options": "a ) a ) 40 , b ) b ) 45 , c ) c ) 36 , d ) d ) 55 , e ) e ) 60", + "Correct": "c", + "Explanation": "\"approach 1 at - least 1 paper back = total - no paper back 8 c 3 - 6 c 3 = 36 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 2 c 1 * 6 c 2 + 2 c 2 * 6 c 1 = 36 answer is c\"" + }, + { + "Answer": 0.13636363636363635, + "Options": "a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22", + "Correct": "d", + "Explanation": "5 c 2 * 3 c 1 = 10 * 3 = 30 12 c 3 = 220 3 / 22 answer : d" + }, + { + "Answer": 0.2857142857142857, + "Options": "a ) 1 / 5 , b ) 2 / 7 , c ) 2 / 9 , d ) 3 / 5 , e ) 4 / 11", + "Correct": "b", + "Explanation": "\"the total number of ways to choose 2 apples is 7 c 2 = 21 the number of ways that include the spoiled apple is 6 c 1 = 6 p ( the spoiled apple is included ) = 6 / 21 = 2 / 7 the answer is b .\"" + }, + { + "Answer": 0.35172413793103446, + "Options": "a ) 51 / 145 , b ) 9 / 25 , c ) 3 / 5 , d ) 94 / 145 , e ) 5 / 6", + "Correct": "a", + "Explanation": "we have 4 shirts , 8 pairs of pants and 18 ties . i . e 30 items total ways of picking two items out of all = 30 c 2 = 435 total ways of picking two items out of all ties = 18 c 2 = 153 required probability = 153 / 435 = 51 / 145 answer : option a" + }, + { + "Answer": 0.038461538461538464, + "Options": "a ) 1 / 13 , b ) 1 / 20 , c ) 1 / 26 , d ) 1 / 50 , e ) 1 / 100", + "Correct": "c", + "Explanation": "there are 13 letters in mediterranean , out of which e appears thrice , n appears twice , a appears twice , and r appears twice . the number of arrangements of mediterranean is therefore 13 ! / 3 ! 2 ! 2 ! 2 ! the number of arrangements in which the first letter is e and the last letter is r is 11 ! / 2 ! 2 ! 2 ! ( we are left with 11 letters out of which e appears twice , n appears twice and a appears twice ) . p = ( 11 ! / 2 ! 2 ! 2 ! ) / ( 13 ! / 3 ! 2 ! 2 ! 2 ! ) = 11 / ! 2 ! 2 ! 2 ! \u2217 3 ! 2 ! 2 ! 2 ! / 13 ! = 3 ! / 12 \u2217 13 = 1 / 26 answer : c ." + }, + { + "Answer": 72, + "Options": "a ) 62 , b ) 70 , c ) 72 , d ) 71 , e ) 76", + "Correct": "c", + "Explanation": "\"first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 9 and for second = 8 so total methods = 9 * 8 = 72 answer : c\"" + }, + { + "Answer": 0.75, + "Options": "a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 2 , d ) 3 / 4 , e ) 7 / 8", + "Correct": "d", + "Explanation": "favorable case = the spade is picked in the third draw or later unfavorable case = the spade is picked in either first draw or in second draw probability = favorable outcomes / total out comes also probability = 1 - ( unfavorable outcomes / total out comes ) unfavorable case : probability of spade picked in first draw = 13 / 52 = 1 / 4 i . e . , favorable probability = 1 - ( 1 / 4 ) = 3 / 4 answer option : d" + }, + { + "Answer": 0.10526315789473684, + "Options": "a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10", + "Correct": "b", + "Explanation": "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability w = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b" + }, + { + "Answer": 0.625, + "Options": "a ) 2 / 3 , b ) 5 / 8 , c ) 3 / 5 , d ) 4 / 7 , e ) 5 / 8", + "Correct": "b", + "Explanation": "\"since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . \u2234 required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b\"" + }, + { + "Answer": 0.375, + "Options": "a ) 0.125 , b ) 0.375 , c ) 0.325 , d ) 0.5 , e ) 0.666", + "Correct": "b", + "Explanation": "total number of ways in which h or t can appear in 3 tosses of coin is = 2 * 2 * 2 = 8 ways for 2 t and 1 th thus probability is = p ( htt ) + p ( tth ) + p ( tht ) = 1 / 8 + 1 / 8 + 1 / 8 = 3 / 8 = . 375 answer : b" + }, + { + "Answer": 16, + "Options": "a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20", + "Correct": "c", + "Explanation": "the total number of ways to choose 3 people from 6 is 6 c 3 = 20 . the number of committees without jane or thomas is 4 c 3 = 4 . there are 20 - 4 = 16 possible committees which include jane and / or thomas . the answer is c ." + }, + { + "Answer": 50, + "Options": "a ) 25.0 % , b ) 37.5 % , c ) 50.0 % , d ) 62.5 % , e ) 75 %", + "Correct": "c", + "Explanation": "p ( chile and not madagascar ) = 0.3 * 0.5 = 0.15 p ( madagascar and not chile ) = 0.5 * 0.7 = 0.35 total probability = 0.15 + 0.35 = 0.5 = 50 % the answer is c ." + }, + { + "Answer": 0.012777777777777779, + "Options": "a ) 1 / 200 , b ) 1 / 100 , c ) 1 / 50 , d ) 1 / 40 , e ) 23 / 1800", + "Correct": "e", + "Explanation": "in how many ways we can select a woman and a man from 30 women and 60 men ? in 30 * 60 = 1800 ways . we have a total of 23 couples so , the probability of selecting a couple is 23 / 1800 = 23 / 1800 . ans - e" + }, + { + "Answer": 120, + "Options": "a ) 60 , b ) 80 , c ) 100 , d ) 120 , e ) 140", + "Correct": "d", + "Explanation": "head ' s sit is fixed . now 5 people is left with 5 seats . 5 people can sit on 5 seats in 5 ! ways or 120 ways . ans is d" + }, + { + "Answer": 0.0021645021645021645, + "Options": "a ) 1 / 35 , b ) 1 / 10 , c ) 1 / 3 , d ) 1 / 462 , e ) 1 / 252", + "Correct": "d", + "Explanation": "\"first , total ways to select for all boys and girls , i . e 12 ! / ( 6 ! * 6 ! ) = 12 * 11 * 10 * 9 * 8 * 7 / 6 * 5 * 4 * 3 * 2 = 924 then there are one two way girls can have all same colors , either white or black . the number of ways in which 6 girls can select 6 white balls = 6 c 6 = 1 the number of ways in which 6 girls can select 6 black balls = 6 c 6 = 1 therefore , total favorable outcomes / total outcomes = 2 / 924 = 1 / 462 d\"" + }, + { + "Answer": 0.02857142857142857, + "Options": "a ) 1 / 35 , b ) 1 / 9 , c ) 1 / 10 , d ) 1 / 20 , e ) 1 / 30", + "Correct": "a", + "Explanation": "\"first , total ways to select for all boys and girls , i . e 8 ! / ( 4 ! * 4 ! ) = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / 4 * 3 * 2 * 1 * 4 * 3 * 2 * 1 = 70 then there are one two way girls can have all same colors , either white or black . the number of ways in which 4 girls can select 4 white balls = 4 c 4 = 1 the number of ways in which 4 girls can select 4 black balls = 4 c 4 = 1 therefore , total favorable outcomes / total outcomes = 2 / 70 = 1 / 35 a\"" + }, + { + "Answer": 0.25, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6", + "Correct": "c", + "Explanation": "\"the total number of ways to choose 2 apples is 8 c 2 = 28 the number of ways that include the spoiled apple is 7 c 1 = 7 p ( the spoiled apple is included ) = 7 / 28 = 1 / 4 the answer is c .\"" + }, + { + "Answer": 0.15060240963855423, + "Options": "a ) 36 / 165 , b ) 25 / 166 , c ) 26 / 165 , d ) 32 / 165 , e ) 33 / 165", + "Correct": "b", + "Explanation": "the number of young boys that will die = 0.02 x 25,000 = 500 the number of young girls that will die = 0.03 x 14,000 = 420 the number of young adults that will die = 0.15 x 16,000 = 2400 the required probability = the number of young boys who will die / the total number of people who will die . = 500 / ( 500 + 420 + 2400 ) = 25 / 166 answer : b" + }, + { + "Answer": 0.10714285714285714, + "Options": "a ) 1 / 5 , b ) 3 / 28 , c ) 1 / 7 , d ) 1 / 14 , e ) 5 / 28", + "Correct": "b", + "Explanation": "\"total number of arrangement is = 8 ! = 40320 favorable event i . e a , b and d can be arranged in 3 ! and the remaining can be arranged in 6 ! since abd can be in any of the six positions . so 3 ! * 6 ! / 8 ! = 3 / 28 answer : b\"" + }, + { + "Answer": 0.08333333333333333, + "Options": "a ) 2 / 3 , b ) 1 / 3 , c ) 2 / 9 , d ) 1 / 24 , e ) 1 / 12", + "Correct": "e", + "Explanation": "probability of first member an english teacher = 3 / 9 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 9 x 2 / 8 = 1 / 12 ( e )" + }, + { + "Answer": 0.5357142857142857, + "Options": "a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14", + "Correct": "d", + "Explanation": "\"the required probability c = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer .\"" + }, + { + "Answer": 40, + "Options": "a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 %", + "Correct": "d", + "Explanation": "\"total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability v = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d\"" + }, + { + "Answer": 0.26666666666666666, + "Options": "a ) 4 / 15 , b ) 3 / 40 , c ) 13 / 40 , d ) 7 / 20 , e ) 13 / 22", + "Correct": "a", + "Explanation": "\"p ( 1 and not 2 ) = 2 / 3 * ( 1 - 3 / 5 ) = 4 / 15 . answer : a .\"" + }, + { + "Answer": 0.03125, + "Options": "a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 8 , d ) 1 / 16 , e ) 1 / 32", + "Correct": "e", + "Explanation": "p ( hhhht ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 the answer is e ." + }, + { + "Answer": 0.08333333333333333, + "Options": "a ) 2 / 3 , b ) 1 / 3 , c ) 2 / 9 , d ) 1 / 12 , e ) 1 / 24", + "Correct": "d", + "Explanation": "\"probability of first member an english teacher = 3 / 9 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 9 x 2 / 8 = 1 / 12 ( d ) answer : d\"" + }, + { + "Answer": 0.4166666666666667, + "Options": "a ) 1 / 6 , b ) 3 , c ) 5 / 12 , d ) 7 / 12 , e ) 9 / 12", + "Correct": "c", + "Explanation": "total number of outcomes possible = 36 probability of total score is a prime number = 15 p ( e ) = 15 / 36 = 5 / 12 answer : c" + }, + { + "Answer": 12, + "Options": "a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 12", + "Correct": "e", + "Explanation": "\"6 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 6 x 2 = 12 e\"" + }, + { + "Answer": 0.10526315789473684, + "Options": "a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10", + "Correct": "b", + "Explanation": "\"method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability e = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b\"" + }, + { + "Answer": 0.6666666666666666, + "Options": "a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1", + "Correct": "d", + "Explanation": "\"answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability r is 2 / 3 . d\"" + }, + { + "Answer": 7, + "Options": "a ) 12 . , b ) 9 . , c ) 8 . , d ) 7 . , e ) 6 .", + "Correct": "d", + "Explanation": "\"7 = 3 blue balls + 12 / 3 answer : d\"" + }, + { + "Answer": 42, + "Options": "a ) 40 , b ) 49 , c ) 41 , d ) 56 , e ) 42", + "Correct": "e", + "Explanation": "\"first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 7 and for second = 6 so total methods = 7 * 6 = 42 answer : e\"" + }, + { + "Answer": 0.75, + "Options": "a ) 3 / 4 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16", + "Correct": "a", + "Explanation": "\"bag consists of 20 marbles , of which 5 are blue , 7 are red remainder are white . so , white = 20 - 5 - 7 = 8 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 7 / 20 + 8 / 20 = 15 / 20 = 3 / 4 hence , answer will be a .\"" + }, + { + "Answer": 0.2857142857142857, + "Options": "a ) 3 / 10 , b ) 2 / 5 , c ) 2 / 7 , d ) 10 / 21 , e ) 5 / 11", + "Correct": "c", + "Explanation": "from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 7 . the answer is c ." + }, + { + "Answer": 84.375, + "Options": "a ) 40 , b ) 46 , c ) 52 , d ) 62 , e ) 85", + "Correct": "e", + "Explanation": "\"the probability of survival for each of the first 3 months of life is 1 - 1 / 4 = 3 / 4 , so of 200 newborn 200 * 3 / 4 * 3 / 4 * 3 / 4 = ~ 85 is expected to survive . answer : e .\"" + }, + { + "Answer": 0.16666666666666666, + "Options": "a ) 1 / 7 , b ) 1 / 6 , c ) 1 / 11 , d ) 1 / 13 , e ) 2 / 9", + "Correct": "b", + "Explanation": "\"number greater than 5 is 6 , so only 1 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 1 / 6 b )\"" + }, + { + "Answer": 0.1388888888888889, + "Options": "a ) 1 / 9 , b ) 5 / 36 , c ) 1 / 3 , d ) 2 / 36 , e ) none of the above", + "Correct": "b", + "Explanation": "we have 5 outcomes that satisfy the question : 1 - 5 2 - 4 3 - 3 4 - 2 5 - 1 the probability is ( 1 / 6 * 1 / 6 ) * 5 = 5 / 36 . i think the answer is b" + }, + { + "Answer": 0.33766233766233766, + "Options": "a ) 4 / 19 , b ) 7 / 19 , c ) 12 / 19 , d ) 21 / 99 , e ) 26 / 77", + "Correct": "e", + "Explanation": "\"solution p ( none is defective ) = 18 c 2 / 22 c 2 = 51 / 77 p ( at least one is defective ) = ( 1 - 51 / 77 ) = 26 / 77 . answer e\"" + }, + { + "Answer": 0.35238095238095235, + "Options": "a ) 4 / 19 , b ) 7 / 19 , c ) 12 / 19 , d ) 21 / 95 , e ) 31 / 88", + "Correct": "e", + "Explanation": "\"solution p ( none is defective ) = 17 c 2 / 21 c 2 = 57 / 88 . p ( at least one is defective ) = ( 1 - 57 / 88 ) = 31 / 88 . answer e\"" + }, + { + "Answer": 0.2, + "Options": "a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 25", + "Correct": "b", + "Explanation": "\"total pairs = 6 c 2 = 15 ; same color pairs = 3 c 1 * 1 c 1 = 3 ; prob = 1 / 5 ans b\"" + }, + { + "Answer": 625, + "Options": "a ) 24 , b ) 120 , c ) 625 , d ) 720 , e ) 1024", + "Correct": "c", + "Explanation": "\"5 choices for each of the 4 questions , thus total of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways . answer : c .\"" + }, + { + "Answer": 0.015479876160990712, + "Options": "a ) 1 / 969 , b ) 2 / 969 , c ) 4 / 969 , d ) 5 / 323 , e ) 21 / 789", + "Correct": "d", + "Explanation": "sample space = no . of ways 4 balls were drawn from urn = 19 c 4 = 3876 no . ways 1 blue balls and 3 red were drawn from bag = 6 c 1 * 5 c 3 = 60 probability = 60 / 3876 = 5 / 323 ans - d" + }, + { + "Answer": 0.23040000000000002, + "Options": "a ) 0.12 , b ) 0.23 , c ) 0.35 , d ) 0.45 , e ) 0.65", + "Correct": "b", + "Explanation": "the probability of winning is 40 % = 40 / 100 = 2 / 5 . the probability of not winning is 60 % = 3 / 5 . p ( wwwnn ) = 5 ! 3 ! 2 ! \u2217 ( 25 ) 3 \u2217 ( 35 ) 2 = 144625 \u2248 23 p ( wwwnn ) = 5 ! 3 ! 2 ! \u2217 ( 25 ) 3 \u2217 ( 35 ) 2 = 144625 \u2248 23 ( we multiply by 5 ! 3 ! 2 ! 5 ! 3 ! 2 ! , because wwwnn scenario can occur in several ways : wwwnn , wwnwn , wnwwn , nwwwn , . . . the # of such cases basically equals to the # of permutations of 5 letters wwwnn , which is 5 ! 3 ! 2 ! 5 ! 3 ! 2 ! ) . answer : b ." + }, + { + "Answer": 0.10714285714285714, + "Options": "a ) 3 / 14 , b ) 5 / 28 , c ) 3 / 28 , d ) 1 / 7 , e ) 9 / 28", + "Correct": "c", + "Explanation": "total number of arrangement is = 8 ! = 40320 favorable event i . e ) a , b and c can be arranged in 3 ! and the remaining can be arranged in 6 ! since ab can be in any of the four positions . so 3 ! * 6 ! / 8 ! = 3 / 28 answer : c" + }, + { + "Answer": 0.6, + "Options": "a ) 1 / 10 , b ) 4 / 9 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3", + "Correct": "d", + "Explanation": "prob reqd = 1 - probab of not having eq numbers prob of not having eq no ' s is = prob of having 3 boys and 1 girl or prob of having 3 girls and 1 boy = ( 1 * 3 / 15 ) + ( 3 / 15 * 1 ) = 2 / 5 reqd prob = 1 - 2 / 5 = 3 / 5 answer : d" + }, + { + "Answer": 0.5555555555555556, + "Options": "a ) 4 / 9 , b ) 5 / 9 , c ) 11 / 18 , d ) 7 / 9 , e ) none of the above", + "Correct": "b", + "Explanation": "probability of getting two same faces : 1 and 1 / 6 for other two faces , probabilities : 5 / 6 and 4 / 6 respectively . the dice combinations can be arranged in 4 ! / 2 ! ways total probability = 1 * 1 / 6 * 5 / 6 * 4 / 6 * 4 ! / 2 ! = 5 / 9 answer : b" + }, + { + "Answer": 0.375, + "Options": "a ) 3 / 8 , b ) 2 / 7 , c ) 1 / 3 , d ) 2 / 5 , e ) 6 / 7", + "Correct": "a", + "Explanation": "number of cases favourable to e = 3 total number of cases = 3 + 5 = 8 p ( e ) = 3 / 8 answer is a" + }, + { + "Answer": 0.1111111111111111, + "Options": "a ) 1 / 7 , b ) 1 / 9 , c ) 2 / 7 , d ) 3 / 7 , e ) 2 / 9", + "Correct": "b", + "Explanation": "one person can select one house out of 3 = \\ inline { \\ color { black } 3 c _ { 1 } } ways = 3 . hence , three persons can select one house out of 3 in 3 x 3 x 3 = 9 . therefore , probability that all thre apply for the same house is 1 / 9 b" + }, + { + "Answer": 0.234375, + "Options": "a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64", + "Correct": "a", + "Explanation": "chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days w = 15 / 64 ans a it is ." + }, + { + "Answer": 25.00000000000001, + "Options": "a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 %", + "Correct": "b", + "Explanation": "\"could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . t = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %\"" + }, + { + "Answer": 0.68, + "Options": "a ) 17 / 26 , b ) 17 / 25 , c ) 17 / 20 , d ) 17 / 22 , e ) 17 / 21", + "Correct": "b", + "Explanation": "the number of exhaustive events = \u2075 \u2070 c \u2081 = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . required probability = 34 / 50 = 17 / 25 . nswer : b" + }, + { + "Answer": 40, + "Options": "a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 %", + "Correct": "d", + "Explanation": "total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability = ( 8 / 20 ) * 100 = 40 % . ans : : d" + }, + { + "Answer": 0.5357142857142857, + "Options": "a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14", + "Correct": "d", + "Explanation": "\"the required probability w = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer .\"" + }, + { + "Answer": 0.28571428571428575, + "Options": "a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 7 , d ) 3 / 11 , e ) 4 / 5", + "Correct": "c", + "Explanation": "let a = event that the husband is selected and b = event that the wife is selected . then , p ( a ) = 1 / 7 and p ( b ) = 1 / 5 \u2234 p ( a \u00af ) = ( 1 - 17 / 5 ) = 6 / 7 and = ( 1 - 1 / 5 ) = 4 / 5 required probability = p [ ( a and not b ) or ( b and not a ) ] = p [ ( a and b \u00af ) or ( b and a \u00af ) ] = p ( a and b \u00af ) + p ( b and a \u00af ) = p ( a ) . p ( b \u00af ) + p ( b ) . p ( a \u00af ) = ( 1 / 7 * 4 / 5 ) + ( 1 / 5 * 6 / 7 ) = 10 / 35 = 2 / 7 c" + }, + { + "Answer": 9, + "Options": "a ) 12 . , b ) 9 . , c ) 8 . , d ) 7 . , e ) 6 .", + "Correct": "b", + "Explanation": "\"9 = 5 blue balls + 20 / 5 answer : b\"" + }, + { + "Answer": 0.75, + "Options": "a ) 1 / 7 , b ) 2 / 4 , c ) 3 / 4 , d ) 1 / 8 , e ) 1 / 16", + "Correct": "c", + "Explanation": "probability that either selects a red marble , p ( either ) = 1 - probability that neither selects a blue marble , p ( neither ) probability that tom selects non blue = 10 c 1 / 20 c 1 = 1 / 2 probability that marry selects non blue = 10 c 1 / 20 c 1 = 1 / 2 = > p ( either ) = 1 - ( 1 / 2 * 1 / 2 ) = 1 - 1 / 4 = 3 / 4 . c )" + }, + { + "Answer": 0.8333333333333334, + "Options": "a ) 5 / 6 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16", + "Correct": "a", + "Explanation": "\"bag consists of 20 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 30 - 5 - 9 = 16 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 30 + 16 / 30 = 25 / 30 = 5 / 6 hence , answer will be a .\"" + }, + { + "Answer": 109.35000000000001, + "Options": "a ) 110 , b ) 146 , c ) 152 , d ) 162 , e ) 170", + "Correct": "a", + "Explanation": "\"the probability of survival for each of the first 3 months of life is 1 - 1 / 10 = 9 / 10 , so of 150 newborn 150 * 9 / 10 * 9 / 10 * 9 / 10 = ~ 110 is expected to survive . answer : a .\"" + }, + { + "Answer": 0.9642857142857143, + "Options": "a ) 30 / 28 , b ) 40 / 28 , c ) 25 / 28 , d ) 27 / 28 , e ) 32 / 28", + "Correct": "d", + "Explanation": "there are 6 consonants and 2 vowels in barkhint . probability that at least one of the selected letters is a consonant = 1 - ( probability of both vowels ) probability of both vowels = 2 c 2 / 8 c 2 = 1 / 28 so , answer is 1 - 1 / 28 = 27 / 28 answer : d" + }, + { + "Answer": 0.234375, + "Options": "a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64", + "Correct": "a", + "Explanation": "\"chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days z = 15 / 64 ans a it is .\"" + }, + { + "Answer": 0.8125, + "Options": "a ) 1 / 16 , b ) 5 / 16 , c ) 2 / 5 , d ) 13 / 16 , e ) 27 / 32", + "Correct": "d", + "Explanation": "fair coin is tossed 5 times . hence total number of outcomes = 2 ^ 5 = 32 . problem asks for probability of getting atleast heads twice . hence if we calculate probability of getting heads exactly once and probability of not getting heads at all and subract it from the total probability of the event which is 1 ( as total probability of certain event will be always 1 ) we can get the probability of getting atleast heads twice . probability of getting exactly one head and no heads = ( number of possible outcomes [ htttt , thttt , tthtt , tttht , tttth , ttttt ] = 6 ) / ( total possible outcomes = 32 ) = > 6 / 32 = 3 / 16 hence probability of getting atleast heads twice = 1 - ( 3 / 16 ) = 13 / 16 = > choice [ d ]" + }, + { + "Answer": 0.0625, + "Options": "a ) 1 / 144 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 16 , e ) 1 / 3", + "Correct": "d", + "Explanation": "\"solution - rolling dices is an independent event . the combinations to get 5 are ( 1,4 ) , ( 4,1 ) , ( 2,3 ) , ( 3,2 ) , and total combinations of both dices is 16 . the probability of getting 5 in first attempt is 4 / 16 = 1 / 4 . probability of getting 5 again in second attempt = ( 1 / 4 ) * ( 1 / 4 ) = 1 / 16 . ans d\"" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these", + "Correct": "b", + "Explanation": "\"explanation : number greater than 4 is 5 & 6 , so only 2 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 2 / 6 = 1 / 3 answer : b\"" + }, + { + "Answer": 0.04000000000000001, + "Options": "a ) 1 / 144 , b ) 1 / 25 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 3", + "Correct": "b", + "Explanation": "solution - rolling dices is an independent event . the combinations to get 6 are ( 1,5 ) , ( 5,1 ) , ( 2,4 ) , ( 4,2 ) , ( 3,3 ) and total combinations of both dices is 25 . the probability of getting 6 in first attempt is 5 / 25 = 1 / 5 . probability of getting 6 again in second attempt = ( 1 / 5 ) * ( 1 / 5 ) = 1 / 25 . ans b" + }, + { + "Answer": 576, + "Options": "a ) 144 , b ) 288 , c ) 12 , d ) 576 , e ) none", + "Correct": "d", + "Explanation": "\"solution : let the arrangement be , b g b g b g b g 4 boys can be seated in 4 ! ways . girl can be seated in 4 ! ways . required number of ways , = 4 ! * 4 ! = 576 . answer : option d\"" + }, + { + "Answer": 0.2, + "Options": "a ) 2 / 9 , b ) 3 / 8 , c ) 2 / 7 , d ) 1 / 6 , e ) 1 / 5", + "Correct": "e", + "Explanation": "\"the total number of ways to choose 2 apples is 10 c 2 = 45 the number of ways that include the spoiled apple is 9 c 1 = 9 p ( the spoiled apple is included ) = 9 / 45 = 1 / 5 the answer is e .\"" + }, + { + "Answer": 0.2222222222222222, + "Options": "a ) 2 / 3 , b ) 2 / 5 , c ) 2 / 7 , d ) 2 / 9 , e ) 2 / 11", + "Correct": "d", + "Explanation": "the total number of ways to choose 2 apples is 9 c 2 = 36 the number of ways that include the spoiled apple is 8 c 1 = 8 p ( the spoiled apple is included ) = 8 / 36 = 2 / 9 the answer is d ." + }, + { + "Answer": 0.4, + "Options": "a ) 1 / 7 , b ) 1 / 8 , c ) 2 / 5 , d ) 3 / 4 , e ) 7 / 8", + "Correct": "c", + "Explanation": "\"good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 1 / 5 of the garden is green , so , since all the flowers must be either green or yellow , we know that 4 / 5 are yellow . we ' re also told there is an equal probability of straight or curved , 1 / 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( straight ) = 4 / 5 * 1 / 2 = 2 / 5 , or c .\"" + }, + { + "Answer": 0.047619047619047616, + "Options": "a ) 5 / 42 , b ) 2 / 42 , c ) 7 / 42 , d ) 2 / 7 , e ) 5 / 7", + "Correct": "b", + "Explanation": "since jill owns 5 of the pen , the subset from which the 2 pens hould be chosen are the 2 pens not owned by jill fom the universe of 7 . the first pen can be one of the 2 from the 7 with probability 2 / 7 . the second pen can be one of the 1 from the 6 remaining with probability 1 / 6 , the total probability will be 2 / 7 \u00d7 1 / 6 . on cancellation , this comes to 2 / 42 . thus , the answer is b - 2 / 42 ." + }, + { + "Answer": 0.30000000000000004, + "Options": "a ) 3 / 10 , b ) 7 / 19 , c ) 12 / 19 , d ) 21 / 95 , e ) none", + "Correct": "a", + "Explanation": "\"solution p ( none is defective ) = 21 c 2 / 25 c 2 = 7 / 10 p ( at least one is defective ) = ( 1 - 7 / 10 ) = 3 / 10 . answer a\"" + }, + { + "Answer": 6, + "Options": "a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 12", + "Correct": "d", + "Explanation": "\"total cases : 12 ( student one has 4 options and student two has three options , 4 x 3 = 12 ) non - favourable cases : 6 ( when two students sit together . students in desk 1 and desk 2 , in desk 2 and desk 3 , in desk 3 and desk 4 ) for each of these cases there are two possibilities because the positions can be interchanged . hence 2 x 3 = 6 . so favourable cases : 12 - 6 = 6 . answer : d\"" + }, + { + "Answer": 0.16666666666666666, + "Options": "a ) 1 / 6 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2", + "Correct": "a", + "Explanation": "\"the number of exhaustive outcomes is 36 . let e be the event of getting doublet on the dies is 6 / 36 = 1 / 6 p ( e ) = 1 / 6 . a )\"" + }, + { + "Answer": 0.421875, + "Options": "a ) 1 / 2 , b ) 3 / 4 , c ) 7 / 8 , d ) 27 / 32 , e ) 27 / 64", + "Correct": "e", + "Explanation": "favorable case = the club is picked in the third draw or later unfavorable cases = the club is picked in either first draw , second draw or third draws probability = favorable outcomes / total out comes also probability = 1 - ( unfavorable outcomes / total out comes ) unfavorable case 1 : probability of club picked in first draw = 13 / 52 = 1 / 4 unfavorable case 2 : probability of club picked in second draw ( i . e . first draw is not club ) = ( 39 / 52 ) * ( 13 / 52 ) = ( 3 / 4 ) * ( 1 / 4 ) = 3 / 16 unfavorable case 3 : probability of club picked in third draw ( i . e . first and 2 nd draws are not clubs ) = ( 39 / 52 ) * ( 39 / 52 ) * ( 13 / 52 ) = ( 3 / 4 ) * ( 3 / 4 ) * ( 1 / 4 ) = 9 / 64 total unfavorable probability = ( 1 / 4 ) + ( 3 / 16 ) + ( 9 / 64 ) = ( 16 / 64 ) + ( 12 / 64 ) + ( 9 / 64 ) = 37 / 64 i . e . , favorable probability = 1 - ( 37 / 64 ) = 27 / 64 answer : option e" + }, + { + "Answer": 0.03125, + "Options": "a ) 3 / 5 , b ) 1 / 2 , c ) 1 / 5 , d ) 1 / 8 , e ) 1 / 32", + "Correct": "e", + "Explanation": "\"on the first three flips , you must get heads . whats the probability of getting heads ? its 1 / 2 so for the first three flips , your probability is ( 1 / 2 ) ^ 3 = 1 / 8 now for the last two , you want to get tails only . whats the prob of getting tails ? well , its the same as prob of getting a heads , namely , 1 / 2 for the last two flips , your probability is ( 1 / 2 ) ^ 2 = 1 / 4 so your overall probability for the event in question is 1 / 8 * 1 / 4 = 1 / 32 answer : e\"" + }, + { + "Answer": 0.08333333333333333, + "Options": "a ) 2 / 3 , b ) 1 / 12 , c ) 2 / 9 , d ) 1 / 2 , e ) 1 / 24", + "Correct": "b", + "Explanation": "\"probability of first member an english teacher = 3 / 9 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 9 x 2 / 8 = 1 / 12 ( b )\"" + }, + { + "Answer": 0.06666666666666667, + "Options": "a ) 1 / 13 , b ) 1 / 14 , c ) 1 / 12 , d ) 1 / 15 , e ) 1 / 16", + "Correct": "d", + "Explanation": "\"taking 2 red shoe the probablity is 3 c 2 from 10 shoes probablity of taking 2 red shoe is 3 c 2 / 10 c 2 = 1 / 15 answer : d\"" + }, + { + "Answer": 0.26666666666666666, + "Options": "a ) 1 / 15 , b ) 2 / 15 , c ) 4 / 15 , d ) 7 / 15 , e ) none of these", + "Correct": "c", + "Explanation": "\"explanation : since every third number starting from 10 will be divisible by 3 , so total number of numbers divisible by 3 are 90 / 3 = 30 numbers which are divisible by 3 and 5 both are numbers which are multiple of 15 . for the range 10 to 99 , 15 is the first number divisible by 15 and 90 is the last number . so total number of numbers divisible by 15 are : ( 90 - 15 ) / 15 + 1 = 5 + 1 = 6 number of numbers which are divisible by 3 are 30 and number of numbers which are divisible by 3 and 5 both are 6 . so number of numbers divisible by 3 and not by 5 are : 30 - 6 = 24 so total probability = 24 / 90 = 4 / 15 answer : c\"" + }, + { + "Answer": 65, + "Options": "a ) a ) 40 , b ) b ) 45 , c ) c ) 50 , d ) d ) 55 , e ) e ) 65", + "Correct": "e", + "Explanation": "approach 1 at - least 1 paper back = total - no paper back 8 c 4 - 5 c 4 = 65 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 3 c 1 * 5 c 3 + 3 c 2 * 5 c 2 + 3 c 3 * 5 c 1 = 65 answer is e" + }, + { + "Answer": 0.98125, + "Options": "a ) 1 , b ) 159 / 160 , c ) 157 / 160 , d ) 4 3 / 160 , e ) 0", + "Correct": "c", + "Explanation": "when trying the first time the probability denise does n ' t pick the correct combination = 3999 / 4000 second time , as the total number of possible combinations reduced by one , not picking the right one would be 3998 / 3999 . third time 3997 / 3998 . . . and the same 75 times . so we get : 3999 / 4000 \u2217 3998 / 3999 \u2217 . . . \u2217 3925 / 39263999 / 4000 \u2217 3998 / 3999 \u2217 . . . \u2217 3925 / 3926 every denominator but the first will cancel out and every nominator but the last will cancel out as well . we ' ll get 3925 / 4000 = 157 / 160 . answer : c ." + }, + { + "Answer": 20, + "Options": "a ) 19 , b ) 20 , c ) 30 , d ) 32 , e ) 25", + "Correct": "b", + "Explanation": "\"first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 5 and for second = 4 so total methods = 5 * 4 = 20 answer : b\"" + }, + { + "Answer": 0.125, + "Options": "a ) 1 / 5 , b ) 3 / 20 , c ) 1 / 8 , d ) 3 / 10 , e ) 1 / 10", + "Correct": "c", + "Explanation": "p ( x is prime ) = 1 / 2 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 1 / 2 * 1 / 4 = 1 / 8 the answer is c ." + }, + { + "Answer": 0.009615384615384616, + "Options": "a ) 1 / 13 , b ) 1 / 15 , c ) 1 / 26 , d ) 1 / 52 , e ) 1 / 104", + "Correct": "e", + "Explanation": "p ( club ) = 1 / 4 p ( red king ) = 1 / 26 p ( club then a red king ) = 1 / 4 * 1 / 26 = 1 / 104 the answer is e ." + }, + { + "Answer": 0.37499999999999994, + "Options": "a ) 3 / 5 , b ) 1 / 2 , c ) 3 / 4 , d ) 3 / 8 , e ) 3 / 6", + "Correct": "d", + "Explanation": "explanation : there are two cases 1 ) he is telling truth that the die reports 6 , its probability = 3 / 4 * 1 / 6 = 1 / 8 2 ) he is telling lie that the die reports 6 , its probability = 1 / 4 * 5 / 6 = 5 / 24 so required probability = ( 1 / 8 ) / ( 1 / 8 ) + ( 5 / 24 ) = ( 1 / 8 ) / ( 1 / 3 ) = 3 / 8 hencer ( d ) is the correct answer answer : d" + }, + { + "Answer": 0.3076923076923077, + "Options": "a ) 2 / 10 , b ) 4 / 13 , c ) 3 / 5 , d ) 9 / 7 , e ) 1 / 4", + "Correct": "b", + "Explanation": "\"explanation : p ( s \u1d1c k ) = p ( s ) + p ( k ) - p ( s \u2229 k ) , where s denotes spade and k denotes king . p ( s \u1d1c k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : b\"" + }, + { + "Answer": 0.5142857142857142, + "Options": "a ) 12 / 29 , b ) 14 / 31 , c ) 16 / 33 , d ) 18 / 35 , e ) 20 / 37", + "Correct": "d", + "Explanation": "\"the total number of ways to choose 4 children from 8 is 8 c 4 = 70 the number of ways to choose 2 boys and 2 girls is 4 c 2 * 4 c 2 = 6 * 6 = 36 p ( 2 boys and 2 girls ) = 36 / 70 = 18 / 35 the answer is d .\"" + }, + { + "Answer": 0.03125, + "Options": "a ) 1 / 8 , b ) 1 / 32 , c ) 1 / 4 , d ) 1 / 2 , e ) 1 / 16", + "Correct": "b", + "Explanation": "( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 32 answer : b" + }, + { + "Answer": 0.3333333333333333, + "Options": "a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these", + "Correct": "b", + "Explanation": "\"explanation : number greater than 4 is 5 & 6 , so only 2 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 2 / 6 = 1 / 3 answer : b\"" + }, + { + "Answer": 0.047619047619047616, + "Options": "a ) 5 / 42 , b ) 2 / 42 , c ) 7 / 42 , d ) 2 / 7 , e ) 5 / 7", + "Correct": "b", + "Explanation": "since jill owns 5 of the pen , the subset from which the 2 pens hould be chosen are the 2 pens not owned by jill fom the universe of 7 . the first pen can be one of the 2 from the 7 with probability 2 / 7 . the second pen can be one of the 1 from the 6 remaining with probability 1 / 6 , the total probability will be 2 / 7 \u00d7 1 / 6 . on cancellation , this comes to 2 / 42 . thus , the answer is b - 2 / 42 ." + }, + { + "Answer": 0.16666666666666666, + "Options": "a ) 1 / 6 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2", + "Correct": "a", + "Explanation": "\"the number of exhaustive outcomes is 36 . let e be the event of getting doublet on the dies is 6 / 36 = 1 / 6 p ( e ) = 1 / 6 . a )\"" + }, + { + "Answer": 0.421875, + "Options": "a ) 1 / 2 , b ) 3 / 4 , c ) 7 / 8 , d ) 27 / 32 , e ) 27 / 64", + "Correct": "e", + "Explanation": "favorable case = the club is picked in the third draw or later unfavorable cases = the club is picked in either first draw , second draw or third draws probability = favorable outcomes / total out comes also probability = 1 - ( unfavorable outcomes / total out comes ) unfavorable case 1 : probability of club picked in first draw = 13 / 52 = 1 / 4 unfavorable case 2 : probability of club picked in second draw ( i . e . first draw is not club ) = ( 39 / 52 ) * ( 13 / 52 ) = ( 3 / 4 ) * ( 1 / 4 ) = 3 / 16 unfavorable case 3 : probability of club picked in third draw ( i . e . first and 2 nd draws are not clubs ) = ( 39 / 52 ) * ( 39 / 52 ) * ( 13 / 52 ) = ( 3 / 4 ) * ( 3 / 4 ) * ( 1 / 4 ) = 9 / 64 total unfavorable probability = ( 1 / 4 ) + ( 3 / 16 ) + ( 9 / 64 ) = ( 16 / 64 ) + ( 12 / 64 ) + ( 9 / 64 ) = 37 / 64 i . e . , favorable probability = 1 - ( 37 / 64 ) = 27 / 64 answer : option e" + }, + { + "Answer": 0.03125, + "Options": "a ) 3 / 5 , b ) 1 / 2 , c ) 1 / 5 , d ) 1 / 8 , e ) 1 / 32", + "Correct": "e", + "Explanation": "\"on the first three flips , you must get heads . whats the probability of getting heads ? its 1 / 2 so for the first three flips , your probability is ( 1 / 2 ) ^ 3 = 1 / 8 now for the last two , you want to get tails only . whats the prob of getting tails ? well , its the same as prob of getting a heads , namely , 1 / 2 for the last two flips , your probability is ( 1 / 2 ) ^ 2 = 1 / 4 so your overall probability for the event in question is 1 / 8 * 1 / 4 = 1 / 32 answer : e\"" + }, + { + "Answer": 0.08333333333333333, + "Options": "a ) 2 / 3 , b ) 1 / 12 , c ) 2 / 9 , d ) 1 / 2 , e ) 1 / 24", + "Correct": "b", + "Explanation": "\"probability of first member an english teacher = 3 / 9 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 9 x 2 / 8 = 1 / 12 ( b )\"" + }, + { + "Answer": 0.98125, + "Options": "a ) 1 , b ) 159 / 160 , c ) 157 / 160 , d ) 4 3 / 160 , e ) 0", + "Correct": "c", + "Explanation": "when trying the first time the probability denise does n ' t pick the correct combination = 3999 / 4000 second time , as the total number of possible combinations reduced by one , not picking the right one would be 3998 / 3999 . third time 3997 / 3998 . . . and the same 75 times . so we get : 3999 / 4000 \u2217 3998 / 3999 \u2217 . . . \u2217 3925 / 39263999 / 4000 \u2217 3998 / 3999 \u2217 . . . \u2217 3925 / 3926 every denominator but the first will cancel out and every nominator but the last will cancel out as well . we ' ll get 3925 / 4000 = 157 / 160 . answer : c ." + }, + { + "Answer": 0.009615384615384616, + "Options": "a ) 1 / 13 , b ) 1 / 15 , c ) 1 / 26 , d ) 1 / 52 , e ) 1 / 104", + "Correct": "e", + "Explanation": "p ( club ) = 1 / 4 p ( red king ) = 1 / 26 p ( club then a red king ) = 1 / 4 * 1 / 26 = 1 / 104 the answer is e ." + }, + { + "Answer": 0.37499999999999994, + "Options": "a ) 3 / 5 , b ) 1 / 2 , c ) 3 / 4 , d ) 3 / 8 , e ) 3 / 6", + "Correct": "d", + "Explanation": "explanation : there are two cases 1 ) he is telling truth that the die reports 6 , its probability = 3 / 4 * 1 / 6 = 1 / 8 2 ) he is telling lie that the die reports 6 , its probability = 1 / 4 * 5 / 6 = 5 / 24 so required probability = ( 1 / 8 ) / ( 1 / 8 ) + ( 5 / 24 ) = ( 1 / 8 ) / ( 1 / 3 ) = 3 / 8 hencer ( d ) is the correct answer answer : d" + }, + { + "Answer": 0.3076923076923077, + "Options": "a ) 2 / 10 , b ) 4 / 13 , c ) 3 / 5 , d ) 9 / 7 , e ) 1 / 4", + "Correct": "b", + "Explanation": "\"explanation : p ( s \u1d1c k ) = p ( s ) + p ( k ) - p ( s \u2229 k ) , where s denotes spade and k denotes king . p ( s \u1d1c k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : b\"" + }, + { + "Answer": 0.03125, + "Options": "a ) 1 / 8 , b ) 1 / 32 , c ) 1 / 4 , d ) 1 / 2 , e ) 1 / 16", + "Correct": "b", + "Explanation": "( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 32 answer : b" + } + ] +} \ No newline at end of file