math-eval/TAL-SCQ5K huggingface dataset:
Problem: A top hat contains $$3$$ red chips and $$2$$ green chips. Chips are drawn randomly, one at a time without replacement, until all $$3$$ of the reds are drawn or until both green chips are drawn. What is the probability that the $$3$$ reds are drawn? 
Options:
aoVal: A
content: $$\dfrac{3}{10}$$ 
aoVal: B
content: $$\dfrac{2}{5}$$ 
aoVal: C
content: $$\dfrac{1}{2}$$ 
aoVal: D
content: $$\dfrac{3}{5}$$ 
aoVal: E
content: $$\dfrac{7}{10}$$ 
Answer: B
Analysis: There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out $$3$$ red chips, $$3$$ red chips, and $$1$$ green chip, $$2$$ green chips, $$2$$ green chips and $$1$$ red chip, and $$2$$ green chips, and $$2$$ red chips.Because order is important in this problem, there are $$1+4+1+3+6=15$$ ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which $$15-5=10$$. Out of the $$10$$ ways to end the game, $$4$$ of them ends with a red chip.  The answer is $$\dfrac{4}{10}=\dfrac{2}{5}$$, or $$\boxed { (\text{B})\dfrac{2}{5}}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: There is a box containing~$7$ chips numbered $1$, $2$, $3$, $4$, $5$, $6$, and $7$. A chip is drawn randomly from the box. What is the probability that the number on the chip is an even number? (adapted from 2015 AMC 8 Problem, Question \#7) 
Options:
aoVal: A
content: $\frac17$ 
aoVal: B
content: $\frac37$ 
aoVal: C
content: $\frac57$ 
aoVal: D
content: $\frac47$ 
aoVal: E
content: $\frac67$ 
Answer: B
Analysis: $2$, $4$, and $6$ are even numbers. Thus, the probability is $\frac37$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: When three line segments with the lengths of $3$, $$4$$, and $$7$$ are used to form a triangle, it is called a/an~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: Random event 
aoVal: B
content: Certain event 
aoVal: C
content: Impossible event 
aoVal: D
content: None of the above 
Answer: C
Analysis: ∵$$3+4=7$$,  ∴ Three line segments measuring $$3$$, $$4$$, and $$7$$ cannot form a triangle,  ∴ Using three line segments measuring $$3$$, $$4$$, and $$7$$ to form a triangle is an impossible event.  So $$\text{C}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Josh started playing soccer at $$14 : 35$$. He stopped playing at $$16 : 05$$. How much time did he spend playing soccer?~\uline{~~~~~~~~~~}~$$\text{h}$$~\uline{~~~~~~~~~~}~$$\min $$ 
Options:
aoVal: A
content: 1，25 
aoVal: B
content: 1，30 
aoVal: C
content: 2，25 
aoVal: D
content: 2，30 
Answer: B
Analysis: 1h 30min 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many three-digit whole numbers have a ones digit equal to the sum of the hundreds digit and the tens digit? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$45$$ 
Answer: D
Analysis: If the ones digit is $$2$$, the $$2$$ numbers are $$202$$ and $$112$$. For each ones digit, the number of possible numbers is the same as the ones digit. In all, there are $$1+2+3+\cdots+8+9 =45$$ numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Four friends, Edwin, Fred, Gary and Howard, were playing together when one of them broke a vase.  The teacher asked: "Who is the culprit?"  Both Edwin and Howard said, "Not me."  Fred said, "Howard broke the vase."  Gary said, "Fred is the culprit."  If only one of four boys was lying,  broke the vase. 
Options:
aoVal: A
content: Edwin 
aoVal: B
content: Fred 
aoVal: C
content: Gary 
aoVal: D
content: Howard 
Answer: B
Analysis: Either Fred or Howard must be lying since what they said did not tally.  Since only one person was lying, Gary was telling the truth i.e, Fred broke the vase. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis
Problem: How many minutes is it from $$22:45$$ today to $$00:35$$ tomorrow? 
Options:
aoVal: A
content: $$90 $$ 
aoVal: B
content: $$ 100 $$ 
aoVal: C
content: $$ 110 $$ 
aoVal: D
content: $$120 $$ 
aoVal: E
content: $$130$$ 
Answer: C
Analysis: It is $$75$$ minutes from $$22:45$$ to midnight and then another $$35$$ minutes from midnight until $$00:35$$. So the required number of minutes is $$75 + 35 = 110$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: The code is a two-digit number. The ones digit is 6 more than the tens digit. How many possible codes are there in total? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$3$$ 
Answer: D
Analysis: 17, 28, 39 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers->Complex Forming Numbers (with special requirements)
Source: 其它
Problem: If the correct time now is $$1:15$$ P.M., and if my clock stopped running three and one-half hours ago, when did it stop running? 
Options:
aoVal: A
content: $$8:45$$ A.M. 
aoVal: B
content: $$9:45$$ A.M. 
aoVal: C
content: $$10:45$$ A.M. 
aoVal: D
content: $$4:45$$ P.M. 
Answer: B
Analysis: One hour ago the time was $$12:15$$ P.M.; two hours ago the time was $$11:15$$ A.M.; three hours ago the time was $$10:15$$ A.M. Three and one-half hours ago the time was $$9:45$$ A.M. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: One morning, a rabbit, a dog, a cat, and a duck went looking for food outside.  The rabbit says: "If I get food, the dog will also get food."  The dog says: "If I get food, the cat will also get food."  The cat says "If I get food, the duck will also get food."  In the evening, they find that all of them tell the truth but only two of them get food.~\uline{~~~~~~~~~~}~and~\uline{~~~~~~~~~~}~don\textquotesingle t get food. 
Options:
aoVal: A
content: The rabbit; the dog 
aoVal: B
content: The dog; the cat 
aoVal: C
content: The cat; the duck 
aoVal: D
content: The cat; the rabbit 
Answer: A
Analysis: We can infer that if the rabbit gets food, then all of the other three would get food; if the dog gets food, then both of the cat and duck would get food. Therefore, only when the rabbit and the dog don\textquotesingle t get food, the cat and the duck would get food. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly? (2014 AMC 8 Problems, Question \#12) 
Options:
aoVal: A
content: $\dfrac{1}{9}$ 
aoVal: B
content: $\dfrac{1}{6}$ 
aoVal: C
content: $\dfrac{1}{4}$ 
aoVal: D
content: $\dfrac{1}{3}$ 
aoVal: E
content: $\dfrac{1}{2}$ 
Answer: B
Analysis: There are $3 \times 2 \times 1 = 6$ ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is (B) $\frac{1}{6}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability
Source: 其它
Problem: Nini has $$6$$ identical balls and she wants to place them into $$3$$ identical baskets. How many different ways can she do so? (The baskets cannot be empty.) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$10$$ 
Answer: B
Analysis: Since the balls and baskets are all identical, the order does not matter.  $$6=1+1+4$$  $$6=1+2+3$$  $$6=2+2+2$$  So, there are $3$ ways to place them. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: The numbers from $1$ to $40$ are written on $40$ pieces of paper, respectively, and put into a box. A piece of paper is chosen at random. Find the probability of choosing a number greater than $20$. 
Options:
aoVal: A
content: $$\frac{1}{2}$$ 
aoVal: B
content: $$\frac{1}{3}$$ 
aoVal: C
content: $$\frac{1}{4}$$ 
aoVal: D
content: $$\frac{2}{5}$$ 
Answer: A
Analysis: Half the numbers from $1$ to $40$ are greater than $20$, Thus, the probability of choosing numbers greater than $20$ is $\dfrac{1}{2}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: How many different cubes are there with three faces coloured red and three faces coloured blue? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: B
Analysis: Consider the case where two opposite faces are coloured red. Whichever of the four remaining faces is also coloured red, the resulting arrangement is equivalent under rotation to a cube with top, bottom and front faces coloured red. Hence, there is only one distinct colouring of a cube consisting of three red and three blue faces with two opposite faces coloured red. Now consider the case where no two opposite faces are coloured red. This is only possible when the three red faces share a common vertex and, however these faces are arranged, the resulting arrangement is equivalent under rotation to a cube with top, front and right-hand faces coloured red. Hence there is also only one distinct colouring of a cube consisting of three red and three blue faces in which no two opposite faces are coloured red. Therefore there are exactly two different colourings of the cube as described in the question. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Problem: Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together? ($2018$ AMC $8$ Problem, Question \#$16$) 
Options:
aoVal: A
content: $$1440$$ 
aoVal: B
content: $$2880$$ 
aoVal: C
content: $$5760$$ 
aoVal: D
content: $$182440$$ 
aoVal: E
content: $$362880$$ 
Answer: C
Analysis: Since the two Arabic books and four Spanish books have to be kept together, respectively, we can treat them both as just one book. That means we\textquotesingle re trying to find the number of ways that you can arrange one Arabic book, one Spanish book, and three German books, which is just $\_5P\_5$. Now we multiply this product by $\_2P\_2\times \_4P\_4$~because there are $\_2P\_2$~ways to arrange just two Arabic books, and $\_4P\_4$~ways to arrange just four Spanish books. Multiplying all these together, we have the answer $C$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50,20,20,5$, and $5$. Let $l$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$? 
Options:
aoVal: A
content: $$-18.5$$ 
aoVal: B
content: $$-13.5$$ 
aoVal: C
content: $$0$$ 
aoVal: D
content: $$13.5$$ 
aoVal: E
content: $$18.5$$ 
Answer: B
Analysis: The formula for expected values is $$ \text { Expected Value }=\sum(\text { Outcome } \cdot \text { Probability }). $$ We have $$ \begin{aligned} t \& =50 \cdot \frac{1}{5}+20 \cdot \frac{1}{5}+20 \cdot \frac{1}{5}+5 \cdot \frac{1}{5}+5 \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =(50+20+20+5+5) \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =100 \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =20, \textbackslash\textbackslash{} s \& =50 \cdot \frac{50}{100}+20 \cdot \frac{20}{100}+20 \cdot \frac{20}{100}+5 \cdot \frac{5}{100}+5 \cdot \frac{5}{100} \textbackslash\textbackslash{} \& =25+4+4+0.25+0.25 \textbackslash\textbackslash{} \& =33.5 . \end{aligned} $$  Therefore, the answer is $t-s=(\mathbf{B})-13.5$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Several distinct positive integers no larger than $$10$$ are listed in a row. After inspecting the row of numbers, Mark concluded with astonishment that in each pair of adjacent numbers one is a divisor of the other. At most how many numbers are in the list? ． 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: D
Analysis: $$6$$, $$3$$, $$9$$, $$1$$, $$4$$, $$8$$, $$2$$, $$10$$, $$5$$. Only $$7$$, which is a prime number and doesn\textquotesingle t have multiples except itself among $$1$$\textasciitilde$$10$$, doesn\textquotesingle t appear. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Operational Problem
Problem: A fair $6$-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number? ． 
Options:
aoVal: A
content: $\dfrac{1}{6}$ 
aoVal: B
content: $\dfrac{5}{12}$ 
aoVal: C
content: $\dfrac{1}{2}$ 
aoVal: D
content: $\dfrac{7}{12}$ 
aoVal: E
content: $\dfrac{5}{6}$ 
Answer: D
Analysis: There are $6\cdot 6=36$ ways to roll the two dice, and $6$ of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $\dfrac{30}{2}=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is $\dfrac{15+6}{36}=\dfrac{21}{36}=\boxed {\left (\text{D}\right )\dfrac{7}{12}}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: There are $$10$$ boys in Pat\textquotesingle s math class. If there are twice as many girls as boys in the class, how many grilss are there in the class? 
Options:
aoVal: A
content: $$50$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$20$$ 
Answer: D
Analysis: girls = $10\times2=20$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Problem: How many different $3-$digit numbers are there with the sum of digits as $15$ ? 
Options:
aoVal: A
content: $$69$$ 
aoVal: B
content: $$78$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$51$$ 
aoVal: E
content: $$49$$ 
Answer: A
Analysis: To divide $15+1+1=17$ into three groups, there are $\_{16}C\_{2}=120$ ways in total.  But, since all digits cannot be larger than $9$, except the situations that don\textquotesingle t meet the requirements, there are $120-\_7C\_2-\_6C\_2\times 2=69$ numbers that match the condition. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Gregor, the mathematician, forms two numbers with the digits $$0$$, $$1$$, $$2$$, $$3$$, $$4$$, and $$5$$. Both numbers have three digits, and each digit is used only once. He adds these two numbers. What is the greatest sum Gregor can get? (adapted from 2012 Math Kangaroo Problem, Level 3 - 4, Question \#19) 
Options:
aoVal: A
content: $$753$$ 
aoVal: B
content: $$861$$ 
aoVal: C
content: $$951$$ 
aoVal: D
content: $$1086$$ 
aoVal: E
content: $$1110$$ 
Answer: C
Analysis: If we want to get the sum as large as possible, the hundred digits for both numbers should be as large as possible. Therefore, they should be $$5$$ and $$4$$. The sum of the hundreds digits is $$9$$. For the same reason, the sum of the tens digits should be $$3+2=5$$ and the sum of the ones digits should be $$1 + 0 = 1$$. Therefore, $$\rm C$$ is the correct answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers
Problem: There are $$10$$ cards in a box, which numbered $$1$$, $$3$$, $$4$$, $$6$$, $$7$$, $$11$$, $$15$$, $$16$$, $$18$$, and $$20$$, respectively. Cathy picks a card from the box randomly. What is the probability the number on the card that Cathy picks is a multiple of $3$? 
Options:
aoVal: A
content: $\frac 1{10}$ 
aoVal: B
content: $\frac 3{10}$ 
aoVal: C
content: $\frac 2{5}$ 
aoVal: D
content: $\frac 1{2}$ 
aoVal: E
content: $\frac 3{5}$ 
Answer: C
Analysis: $3, 6, 15,$ and $18$ are multiples of $3$. Thus, the probability is $\frac 2{5}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: There are $$29$$ students in a certain class. $$12$$ of the students have a sister and $$18$$ of the students have a brother. In this class, only Tania, Barbara, and Anna do not have any siblings. How many students from this class have both a brother and a sister? ($$2003$$ Math Kangaroo Problems, Level $$3-4$$, Question \#$$17$$) 
Options:
aoVal: A
content: $$$$None 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: $(12+18)-(29-3)=4$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: In how many ways can the digits in $3433256337$ be rearranged so that two or more $3$s are not adjacent to each other? 
Options:
aoVal: A
content: $$1800$$ 
aoVal: B
content: $$1200$$ 
aoVal: C
content: $$1000$$ 
aoVal: D
content: $$720$$ 
aoVal: E
content: $$120$$ 
Answer: D
Analysis: There are $5$ $3$s in total.  There are $\_5P\_5$ ways for us to arrange the other $5$ letters\textquotesingle{} positions. Then, we can put the five $3$s in the $6$ intervals.  So the answer is $\_5P\_5 \times \_6C\_5=720$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: $$31$$ children stand in a row from the shortest to the tallest. The last child is $$131$$ cm height. Tom stands the $${{10}^{\text{th}}}$$ place counting from the first. No two children have the same height. What is the maximum height of Tom? (The height of all the children are whole numbers.) 
Options:
aoVal: A
content: $$100$$ cm 
aoVal: B
content: $$105$$ cm 
aoVal: C
content: $$110$$ cm 
aoVal: D
content: $$111$$ cm 
aoVal: E
content: $$125$$ cm 
Answer: C
Analysis: $$131-21=110$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Extreme Value in Inclusion-Exclusion for Multi-sets
Problem: In Bill\textquotesingle s class, every student can play either piano, or violin, or both. $24$ of his classmates can play piano, and the number of students that can play violin is half of that of playing piano. If there are $4$ students who can play both, how many students are there in the class? 
Options:
aoVal: A
content: $$68$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$38$$ 
aoVal: D
content: $$34$$ 
aoVal: E
content: $$32$$ 
Answer: E
Analysis: $24+24\div2-4=32$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Source: 其它
Problem: Four students asked their teacher, Mr. Carter, to line up with them to take a picture.  ①If Mr. Carter does not want to stand on either end, how many different ways can they line up for the picture?  ②If Mr. Carter insists on standing on one of the $2$ ends, how many different ways can they line up for the picture? 
Options:
aoVal: A
content: $72$ , $24$ 
aoVal: B
content: $96$ , $24$ 
aoVal: C
content: $72$ , $48$ 
aoVal: D
content: $96$ , $48$ 
Answer: C
Analysis: ①$$3\times 4\times 3\times 2\times 1=72$$ ,  ②$$2\times 4\times 3\times 2\times 1=48$$ . 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Queuing Problems
Problem: $$13\times 21+26\times 23+39\times 11=$$ 
Options:
aoVal: A
content: $$130$$ 
aoVal: B
content: $$1300$$ 
aoVal: C
content: $$260$$ 
aoVal: D
content: $$2600$$ 
Answer: B
Analysis: $$=13\times(21+46+33)=1300$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Source: 其它
Problem: There are 70 beads.~ $$A$$ and $$B$$ take turns taking beads from the pile ($$A$$ goes first). Each person can take between 1 and 4 beads per turn. The person who takes the last bead loses. If $$A$$ wants to win, how many beads should he take on the first turn? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$A$$ can\textquotesingle t win 
Answer: A
Analysis: $$(70-1)\div (1+4)=13\cdots 4$$  $$A$$ takes four bead first, and then regardless of how many beads $$B$$ takes, as long as the sum of $$A$$ and $$B$$ is 5, $$A$$ will win. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: The password for a bankbook has six digits (digits can be $$0$$). Mr. Wang forgets the first digit. The probability that he can get the first digit right on his first try is ． 
Options:
aoVal: A
content: $$\frac{1}{6}$$ 
aoVal: B
content: $$\frac{1}{9}$$ 
aoVal: C
content: $$\frac{1}{10}$$ 
aoVal: D
content: $$\frac{1}{12}$$ 
Answer: C
Analysis: ∵ There are $$10$$ possible situations. However, there\textquotesingle s only one time for him to get the first digit right at the first try.  ∴ The probability that he can get the first digit right at the first try is $$\frac{1}{10}$$.  So $$\text{D}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Statistical Graphs
Problem: Kevin has $3$ regular dice. Each dice has numbers from $1$ to $6$. Which of the following could not be the sum of the numbers on top of the $3$ dice? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$22$$ 
aoVal: E
content: All the above numbers are possible sum 
Answer: D
Analysis: maximum is 6+6+6 = 18. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: $56+127+34+73=$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$190$$ 
aoVal: B
content: $$290$$ 
aoVal: C
content: $$390$$ 
aoVal: D
content: $$490$$ 
Answer: B
Analysis: $56+127+34+73=(56+34)+(127+73)=90+200=290$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: A film on TV lasts $$2$$ hours and $$28$$ minutes, and finishes at $$11.18 \text{p}.\text{m}.$$ At what time does it start? 
Options:
aoVal: A
content: $$8.10\text{p}.\text{m}.$$ 
aoVal: B
content: $$8.50\text{p}.\text{m}.$$ 
aoVal: C
content: $$9.10\text{p}.\text{m}.$$ 
aoVal: D
content: $$9.50\text{p}.\text{m}.$$ 
aoVal: E
content: $$1.46\text{p}.\text{m}.$$ 
Answer: B
Analysis: $$8.50\text{p}.\text{m}.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation