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Question
Solve the following:
A and B throw a die alternatively till one of them gets a 3 and wins the game. Find the respective probabilities of winning. (Assuming A begins the game)
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Solution
Let A ≡ the event that player A gets 3
B ≡ the event that player B gets 3
When a die is thrown, the probability of getting 3 is `1/6`
∴ P(A) = `1/6`, P(A') = 1 – P(A) = `5/6`
Similarly, P(B) = `1/6`, P(B') = `5/6`
It is given that A starts the game and he will win in the following mutually exclusive ways.
(i) A happens i.e., A wins at the first throw.
(ii) A' n B' n A happens i.e., A wins in the third throw when A and B fail in the first and second throws.
(iii) A' n B' n A' n B' n A happens i.e., A wins at the fifth throw when A and B fail at the 1st, 2nd, 3rd, 4th throw, and so on.
∴ P(A wins)= P(A) + P(A' ∩ B' ∩ A) + P(A' ∩ B' ∩ A' ∩ B' ∩ A) + .......
= P(A) + P(A') · P(B') · P(A) + P(A') · P(B') · P(A') · P(B') · P(A) + .......
= `1/6 + 5/6 xx 5/6 + 1/6 + 5/6 xx 5/6 xx 5/6 xx 5/6 xx 1/6` + ...
= `1/6[1 + (5/6)^2 + (5/6)^4 + ...]`
= `1/6[1/(1 - (5/6)^2)]` ...[it is an infinite geometric series with a = 1, r = `25/36`]
= `1/6[1/((11/36))]`
= `6/11`
P(B wins)= 1 – P(A wins)
= `1 - 6/11 = 5/11`
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