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Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5? | Here S = {1, 2, 3, 4, ...., 19, 20}
Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E)/n(S) = 9/20. |
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? | Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2
= (7 x 6) / (2 x 1) = 21.
Let E = Event of drawing 2 balls, none of which is blue.
n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls.
= 5C2
= (5 x 4) / (2 x 1) = 10.
P(E) = ... |
In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green? | Total number of balls = (8 + 7 + 6) = 21.
Let E = event that the ball drawn is neither red nor green, which is the event that the ball drawn is blue.
n(E) = 7.
P(E) = n(E) / n(S) = 7 / 21 = 1 / 3. |
What is the probability of getting a sum 9 from two throws of a dice? | In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum = {(3, 6), (4, 5), (5, 4), (6, 3)}.
P(E) = n(E) / n(S) = 4 / 36 = 1 / 9. |
Three unbiased coins are tossed. What is the probability of getting at most two heads? | 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. |
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even? | In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.
n(E) = 27.
P(E) = n(E) ... |
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: | Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number of ways of selecting 3 students out of 25
= 25C3
= (25 x 24 x 23) / (3 x 2 x 1) = 2300.
n(E) = (10C1 x 15C2)
= 10 x (15 x 14) / (2 x 1) = 1050.
P(E) = n(E) / n(S) = 1050 / 2300 = 21 / 46. |
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