Question

Problem 10 (10 points). High school A and B play a series of local basketball championship games until one of them wins three games. Suppose that the games are independent and the probability that High school A wins a game is 0.60. (1) Find the probability that the series ends in live games. (2) If the series ends in five games, what is the probability that School A wins?

Answer 1

P(A win) = 0.6

P(B win) = 1 - 0.6 = 0.4

1) P(series ends in 5 games) = P(A wins 3 games) + P(B wins 3 games)

As it says until one of them wins three games hence the 5th game will win by the team who is winning the series.

P(A wins 3 games) = P(A wins 2 in first 4 games) * P(A wins in 5th)

= $\binom{4}{2}*0.6^2 *0.4^2 *0.6 = 0.20736$

(4, 2) used to find number of ways to win 2 games in first 4 games, 0.6^2 is probability of A's winning 2 games and 0.4^2 is probability of B's winning two games. and last 0.6 is there because A is winning 5th game.

Same way for P(B wins 3 games) = P(B wins 2 in first 4 games) * P(B wins in 5th)

= $\binom{4}{2}*0.4^2 *0.6^2 *0.4 = 0.13824$

Same explanation just in last we got 0.4 because P(B wins) = 0.4

So P(series ends in 5 games) = 0.20736 + 0.13824 = 0.3456

2) P(A wins | series ends in 5 games) = $\frac{0.20736}{0.3456}=0.6$

Hence 0.6 is the probability that School A wins if the series ends in 5 games.

Please comment if any doubt. Thank you.