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Question

 $$A$$ and $$B$$ play a match in which the chances of their winning a game is $$\frac{1}{4}$$ for both of them and $$\frac{1}{2}$$ is the probability that match is being drawn. Match is finished as soon as either player wins two games. Find the probability that the match will be finished in $$4$$ or less games.


A
916
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B
716
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C
12
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D
14
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Solution

The correct option is B $$\dfrac{1}{2}$$
We have to find the probability that match will be finished in $$4$$ or less games.
We will  find the ways in which match is not finished in $$4$$ games.
There are following $$4$$ mutually exclusive ways:
(1) All the games are drawn.
(2) $$A$$ and $$B$$ both win one game each and remaining two are drawn.
(3) $$A$$ wins one game and remaining three are drawn.
(4) $$B$$ wins one game and remaining three are drawn. 
So the probability that match is not finished in 4 games is equal to the sum of probabilities of $$4$$ ways.
$$=\displaystyle  \left ( \frac{1}{2} \right )^{4}+^{4}C_{2}\left ( \frac{1}{4} \right )^{2}\left ( \frac{1}{2} \right )^{2}+^{4}C_{1}\left ( \frac{1}{4} \right )\left ( \frac{1}{2} \right )^{3}+^{4}C_{1}\left ( \frac{1}{4} \right )\left ( \frac{1}{2} \right )^{3}$$
$$\displaystyle =\frac{1}{16}+\frac{3}{16}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$
Hence the required probability $$=\displaystyle 1-\frac{1}{2}=\frac{1}{2}$$

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