Math
posted by Urgent!! on .
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.

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