Question

Teams A and Bare in a seven-game playoff series; the team that wins four games is the team that wins the series. Assume that both teams are evenly matched (i.e., the probability of winning each game is 50/50). (1) Team A won the first two games. What is the probability that team B will win the series? (2) continue to assume that Team A has already won two games, but the teams are not evenly matched. Assume that B is a better team. It's better in that its probability of beating Team A in any one game is .60. What is the probability that Team B will win the series?

Answer 1

Solution:

Given,

total number of games = 7

a team has to win 4 games or more out of 7 games to win the series

assuming that team had won first two games ,

so for winning team B has to win 4 games or more out of 5 games

binomial probability distribution

Formula:

P_{(k out of n )}= n!*p^k * q^n-k / k! *(n - k)!

( 1 )

n = 5

p = 0.5

q = 0.5

P( x $\geq$ 4 ) = 5!*0.5⁴ * 0.5^5-4 / 4! *(5 - 4)! + 5!*0.5⁵ * 0.5^5-5 / 5! *(5 - 5)!

= 0.1563 + 0.0313

= 0.1875

( 2 )

n = 5

p = 0.6

q = 0.4

P( x $\geq$ 4 ) = 5!*0.6⁴ * 0.4^5-4 / 4! *(5 - 4)! + 5!*0.6⁵ * 0.4^5-5 / 5! *(5 - 5)!

= 0.2592 + 0.0778

= 0.3370