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A chess game between Kamsky and Anand is...

A chess game between Kamsky and Anand is won by whoever first wins 2 out of 3 games. Kamsky's chance of winning, drawing or lossing a particular game are `p,q,r`. The games are independent and `p+q +r=1`. Prove that the probability that Kamsky wins the match is `(p^2(P+3r))/((p+r)^3)`.

A

`na^2b^(n-1)`

B

`na^2b^(n-2)(b+(n-1)c)`

C

`na^2bc^(n-1)`

D

`nab^(n-1)(b+nc)`

Text Solution

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The correct Answer is:
(b)
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