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Question

$$A$$ and $$B$$ play a match to be decided as soon as either has won two games. The chance of either winning a game is $$\displaystyle \dfrac{1}{20}$$ and of its being drawn $$\displaystyle \dfrac{9}{10}.$$ What is the the chance that the match is finished in $$10$$ or less games?


A
0.18 approx.
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B
0.16 approx.
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C
0.17 approx.
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D
0.15 approx.
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Solution

The correct option is B $$\displaystyle 0.17 $$ approx.
If the match is not finished in $$10$$ games, either of the following events. will possibly occur : 
(i) All the games are drawn 
(ii) $$A$$ and $$B$$ each win one game and the rest $$8$$ games are drawn. 
(iii) $$A$$ or $$B$$ wins one game and the rest $$9$$ games are drawn. The corresponding probabilities of the above events are
    (i) $$\displaystyle \left ( \frac{9}{10} \right )^{10}, $$ (ii) 2.$$\displaystyle ^{10}C_{2}\left ( \frac{1}{20} \right )^{2}\left ( \frac{9}{10} \right )^{8}, $$ (iii) 2.$$\displaystyle ^{10}C_{1}\left ( \frac{1}{20} \right )\left ( \frac{9}{10} \right )^{9},$$
Since these events are mutually exclusive, the probability of
the game not being finished in 10 games $$\displaystyle  =\left ( \frac{9}{10} \right )^{10}+2^{10}C_{2}\left ( \frac{1}{20} \right )^{2}\left ( \frac{9}{10} \right )^{8}+2^{10}C_{1}\left ( \frac{1}{20} \right )\left ( \frac{9}{10} \right )^{9}$$ $$\displaystyle  =\frac{9^{8}\times 387}{2\times 10^{10}}$$
Hence the probability that the match is finished in $$10$$ or less number of games $$\displaystyle =1-\frac{9^{8}\times 387}{2x\times 10^{10}}=0.17 $$ approx.

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