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In a set of five games in tennis between two players A and B,the probability of a player winning a game is $\frac{2}{3}$ who has won the earlier game.A wins the first game.What is the probability that A will win atleast three of the next four games?


Since the probability of $A$ winning atleast three of the next four games is asked.This is possible if $A$ wins 2nd,3rd,4th games or 2nd,3rd,5th games or 3rd,4th,5th games or 2nd,4th,5th games or 2nd,3rd,4th,5th games.

Let the probability of a player winning a game is $p$ if he has not won the earlier game.If he has won the earlier game,then the probability of a player winning a game is $\frac{2}{3}$ as given in the question.

So the required probability is $\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}+\frac{2}{3}\times\frac{2}{3}\times p+p\times\frac{2}{3}\times\frac{2}{3}+\frac{2}{3}\times p \times \frac{2}{3}+\frac{2}{3}\times\frac{2}{3}\times \frac{2}{3}\times\frac{2}{3}$

but i dont know what is the value of $p$,since it is not given in the question.I put $p=\frac{1}{2}$ but it is giving wrong answer.

Correct answer is $\frac{4}{9}$,i dont know where have i made the mistake or my logic is not correct?Please help me.Thanks.

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$p$ is just the probability that the other player doesn't win, which is 1/3.

Also you need to multiply all the terms except the last by another 1/3, since B is winning one game, the probability of which is 1/3.

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