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Question
Solve the following:
Consider independent trails consisting of rolling a pair of fair dice, over and over What is the probability that a sum of 5 appears before sum of 7?
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Solution
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let event A: The sum is 5 in a trial.
A = {(2, 3), (3, 2), (1, 4), (4, 1)}
∴ P(A) = `4/36 = 1/9`
Let event B: The sum is 7 in a trial.
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}
∴ P(B) = `6/36 = 1/6`
Let event C: Neither sum is 5 nor 7.
P(C) = 1 – P(A) – P(B)
= `1 - 1/9 - 1/6`
= `26/36`
Let the sum of 5 appear in nth trial for the first time and the sum of 7 has not occurred in first (n – 1) trials.
Probability of this event = [P(C)]n–1 P(A)
Required probability = `sum_("n" = 1)^oo (26/36)^("n" - 1) (1/9)`
= `1/9(1/(1 - 26/36))`
= `1/9(18/5)`
= `2/5`.
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