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Question
A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins
Solution
Total of 7 on the dice can be obtained in the following ways:
(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)
Probability of getting a total of 7 = `6/36=1/6`
Probability of not getting a total of 7 = `1-1/6=5/6`
Total of 10 on the dice can be obtained in the following ways:
(4, 6), (6, 4), (5, 5)
Probability of getting a total of 10 = `3/36=1/12`
Probability of not getting a total of 10 = `1-1/12=11/12`
Let E and F be the two events, defined as follows:
E = Getting a total of 7 in a single throw of a dice
F = Getting a total of 10 in a single throw of a dice
`P(E)=1/6, P(barE)=5/6P(F)=1/12, P(barF)=11/12`
A wins if he gets a total of 7 in 1st, 3rd or 5th ... throws
Probability of A getting a total of 7 in the 1st throw = `1/6`
A will get the 3rd throw if he fails in the 1st throw and B fails in the 2nd throw.
Probability of A getting a total of 7 in the 3rd throw = `P(barE)P(barF)P(barE)=5/6xx11/12xx1/6`
Similarly, probability of getting a total of 7 in the 5th throw = `P(barE)P(barF)P(barE)P(barF)P(E)=5/6xx11/12xx5/6xx11/12xx1/6 " an so on"`
Probability of winning of A = `1/6+(5/6xx11/12xx1/6)+(5/6xx11/12xx5/6xx11/12xx1/6)+...=(1/6)/(1-5/6xx11/12)=12/17`
∴ Probability of winning of B = 1 − Probability of winning of A =`1-12/17=5/17`
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