A pair of fair dice is rolled until a sum of either 5 or 7 is obtained. Then, the probability that 5 comes before 7 is .…

Solution:

Given a pair of dice is rolled until either 5 or 7 is obtained

The total probability in the sample space is 6² = 36

Let A be the event that the sum 5 occurs

Let B be the event that the sum 7 occurs

C be the event that neither the sum 5 nor sum 7 occurs

Therefore, P(A) = 4/36, P(B) = 6/36, P(C) = 26/36

The probability that A occurs before B = P(A) + P(C).P(A) + P(C)P(C)P(A) +….

C = P(A) + P(C)P(A) + P(C)²P(A) + …

Using GP, we get = P(A) /[1 - P(C) ] = [(4/36)]/[1 - (26/36)] = 2/5.

A pair of fair dice is rolled until a sum of either 5 or 7 is obtained. Then, the probability that 5 comes before 7 is .…

Summary:

A pair of fair dice is rolled until a sum of either 5 or 7 is obtained. Then, the probability that 5 comes before 7 is 2/5.