  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.  B
0.16 approx.  C
0.17 approx.  D
0.15 approx.  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.Maths

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