Posted by **Urgent!!** on Wednesday, February 13, 2013 at 8:49pm.

In a World Series, two teams play each other in at least four and at most seven games. The first team to win four games is the winner of the World Series. Assuming that both teams are equally matched, what is the probability that a World Series will be one (a) in four games? (b) in five games? (c) in six games? (d) in seven games? Explain.

- Math -
**Reiny**, Wednesday, February 13, 2013 at 9:53pm
Be "equally matched" I will assume

P(win) = 1/2 = prob(loss)

winning in 4 games = (1/2)^4 = 1/16

in 5 games, has to lose once)

LWWWW

WLWWW

WWLWW

WWWLW ----- each of these has a prob of (1/2)^5

but there are 4 cases,

so prob(5games) = 4/32 = 1/8

6 games , 2 losses, 4 wins

number of ways = 6!/(2!4!) = 15 , but that includes the case of ending with a loss, which can't happen

so number of 6 games is 14

prob(6games) = 14/(1/2)^6 = 14/64 = 7/32

7 games, 4 wins, 3 losses

number of cases = 7!/(3!4!) = 35 , less the case of ending in L

we have 34 such cases

prob(7 games) = 34/(1/2)^7 = 34/128 = 17/64

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