Suppose the probability for A to win a game against B is 0.4. If A has an option of playing either a “best of 3 games'' or a “best of 5 games match against B, which option should be chosen so that the probability of his winning the match is higher? (No game ends in a draw.)

Answer

A, B, C

Explanation

Case I When A plays 3 games against B.
In this case, we have n=3, p=0.4 and q=0.6
Let X denote the number of wins. Then,
`P(X=r)=overset(3)""C_(r )(0.4)^(r )(0.6)^(3-r)p,r=0,1,2,3`
`:. P_(1)`=probability of winning the best of 3 games
`=P(Xge2)`
`=P(X=2)+P(X=3)`
`=overset(3)""C_(2)(0.4)^(2)(0.6)^(1)+overset(3)""C_(3)(0.4)^(3)(0.6)^(0)`
`=0.288+0.064=0.352`
Case II When A plays 5 games against B.
In this case, we have
n=5,p=0.4 and q=0.6
Let X denotes the number of wins in 5 games.
Then,
`P(X=r)=overset(5)""C_(r )(0.4)^(r )(0.6)^(5-r), "where" r=0,1,2,...,5`
`:. P_(2)`=probability of winning the best of 5 games
`=P(Xge3)`
`=P(X=3)+P(X=4)+P(X=5)`
`=overset(5)""C_(3)(0.4)^(3)(0.6)^(2)+overset(5)""C_(4)(0.4)^(4)(0.6)+overset(5)""C_(5)(0.4)^(5)(0.6^(0)`
`=0.2304+0.0768+0.1024=0.31744`
Clearly, `P_(1)gtP_(2)`. Theerefore, first option i.e. 'best of 3 games' has higher probability of winning the match.