Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the teams is stronger than the other and wins each game with probability .6 independently of the outcomes of the other games. Find that probability that the stronger team wins the series.
+6248 \(\text{The probability the stronger team wins is the series, }\\ \text{is the sum of the probabilities that they win 3 games out of 3,4,5,6}\\ \text{times the probability they get their fourth win}\\ p = \left(\sum \limits_{n=3}^6 \dbinom{n}{3}(0.6)^3 (0.4)^{n-3}\right)(0.6) = \dfrac{11097}{15625} \approx 0.71\)
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