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 the probability, for i = 4, 5, 6, 7, that the stronger team wins the series in exactly i games. Compare the probability that the stronger team wins with the probability that it would win a 2-out of-3 series.

Two athletic

hansel

Answered by Aaron 2 years ago

First the probability of winning in 4 games is given by,

$$(0.6)^4$$

now for *i = 5,6,7* (look up negative binomial),

$$P(\text{win in i games})=(C(i,4)-C(i-1,4))(0.6)^4(1-0.6)^{i-4}=C(i-1,3)(0.6)^4(0.4)^{i-4}$$

and for 2 out of 3,

$$P(\text{2 out of 3})=(C(3,2)-C(2,2))(0.6)^2(0.4)^1$$

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