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Two players A and B play a match which consists of a series of games (independent). Whoever first wins two games not necessarily consecutive, wins the match. The probability of A’s winning, drawing or losing a game against B is $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$ respectively. It is known that A won the match at the end of 11th game, the probability that B wins only one game is

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$\frac{10}{11}$

$\frac{8}{11}$

$\frac{3}{11}$

$\frac{9}{11}$

answer is C.

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Detailed Solution

E₁ – A wins the match at end of 11^th game
E₂ – B Wins exactly one game

$∴ P (\frac{E_{2}}{E_{1}}) = \frac{9}{11}$

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