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.)

Explanation

The probability `p_(1)` of winning the best of three games is equal to the sum of the probability of swimming two games and the probability of swining three gamws, which is given by
`p_(1)=""^(3)C_(2)(0.6)(0.4)^(2)+""^(3)C_(3)(0.4)^(3)`
`=0.288+0.064=0.352` [Using binomial distribution]
Similarly, the probability of swinning the best five games is
`p_(2)=` Probability of winning three games
+ Probability of swinning four games
+ Probability of swining 5 gmes
`=""^(5)C_(3)(0.6)^(2)(0.4)^(3)+""^(5)C_(4)(0.4)^(4)+""^(5)C_(5)(0.4)^(5)`
`=0.2304+0.0768+0.01024=0.31744`
As `p_(1)gtp_(2),` A must choose the first offer, i.e., best of three games.