Solution

8 can be thrown with a pair of dice in the following ways (2,6), (6,2), (4,4), (5,3), (3,5) So, Probability of throwing ‘8’ = 5/36 And Probability of not throwing ‘8’ = 31/36 And 7 can be thrown with a pair of dice in the following ways (1,6),  (6,1), (2,5), (5,2), (3,4), (4,3) So, the probability of throwing a ‘7’ = 6/36=  1/6 And Probability of not throwing ‘7’ = 5/6 Let E1 be the event of the throwing a ‘8’ in a single throw of a pair of dice and E2 be the event of throwing a ‘7’ in a single throw of a pair of Dice Then P(E1) = 5/36 , P(E2) = 1/6 And P(¯E1)= 31/36, P(¯E2) = 5/6 Tom wins if he throws ‘8’ in first or third or fifth............ throws. Probability of Tom throwing a ‘8’ in First throw = P(E1) = 5/36 And Probability of Tom throwing a ‘8’ in third throw = P(¯E1 ∩ ¯E2  ∩ E1 ) = P(¯E1) × P(¯E2) × P(E1) = 31/36  ×  5/6  ×5/36  Similarly, Probability of Tom throwing a ‘8’ in fifth throw = P(¯E1) × P(¯E2) × P(¯E1) × P(¯E2) × P(E1) = (31/36)^2×(5/6)² ×5/36 Hence, Probability of winning of Tom = P[E1∪(¯E1 ∩ ¯E2  ∩ E1)∪ (¯E1∩ ¯E2  ∩ ¯E1∩¯E2∩ E1)∪……. ] = P[E1+(¯E1 ∩ ¯E2  ∩ E1)+ (¯E1∩ ¯E2  ∩ ¯E1∩¯E2∩ E1)+⋯…. ] Sum of n terms of Geometric Progression , if r < 1 = a/(1-r) ∴ Probability of winning of Tom = (5/36)/(1-(31/36  × 5/6)) = 30/61 Thus, Probability of winning of Jerry = 1 - 30/61 = 31/61