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 $\frac{1}{20}$ and of its being drawn $\frac{9}{10}.$ 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.

Answer 1

Answer: (C)

$0.17$ 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) ${\left(\frac{9}{10}\right)}^{10},$ (ii) 2.${}^{10}{C}_2{\left(\frac{1}{20}\right)}^2{\left(\frac{9}{10}\right)}^8,$ (iii) 2.${}^{10}{C}_1\left(\frac{1}{20}\right){\left(\frac{9}{10}\right)}^9,$
Since these events are mutually exclusive, the probability of
the game not being finished in 10 games $={\left(\frac{9}{10}\right)}^{10}+2^{10}{C}_2{\left(\frac{1}{20}\right)}^2{\left(\frac{9}{10}\right)}^8+2^{10}{C}_1\left(\frac{1}{20}\right){\left(\frac{9}{10}\right)}^9$ $=\frac{9^8\times 387}{2\times {10}^{10}}$
Hence the probability that the match is finished in $10$ or less number of games $=1-\frac{9^8\times 387}{2x\times {10}^{10}}=0.17$ approx.