Question

In an NBA (National Basketball Association) championship series, the team that wins four games out of seven is the winner. Suppose that teams A and B face each other in the championship games and that team A has probability 0.55 of winning a game over team B.

(a) What is the probability that team A will win the series in 5 games?

(b) What is the probability that team A will win the series?

(c) What is the probability that team B will win the series in 6 games?

Answer 1

Solution

Given  P(winning)=0.55

(a) What is the probability that team A will win the series in 5 games

Let X =number of losing precesde 4 wins 

If A win the series in 5 game 

\( \implies P(X=1)=C_4^3\left(0.55\right)^4\left(0.45\right) \)

                             \( =2^2\times \left(0,55\right)^4\times \:\left(0.45\right) \)

                             \( =0.16 \)

Therefore. \( P(X\leq 1)=0.16 \)

(b) What is the probability that team A will win the series

\( \implies P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) \)

                             \( =0.251+\left(0.55\right)^4\left(10\left(0.45\right)^2+20\left(0.45\right)^3\right) \)

                             \( =0.251+0.35=0.6 \)

Therefore. \( P(X\leq 3)=0.6 \)

(c) What is the probability that team B will win the series in 6 games

Let Y be the number of losing that precede 4 wins for team B. P(winning)=0.45

\( \implies P(X\leq2)=\sum _{n=0}^2\:C_{x+3}^3\left(0.45\right)^4\left(0.55\right)^x \)

                             \( =\left(0.45\right)^4+4\left(0.45\right)^4\left(0.55\right)+10\left(0.45\right)^4\left(0.55\right)^2 \)

                             \( =\left(0.45\right)^4\left(4\times 0.55+10\left(0.55\right)^2\right) \)

                             \( =0.041\left(2.2+3.025\right)=0.214 \)

Therefore.  \( P(X\leq 2)=0.214 \)

Therefore.

a). \( P(X\leq 1)=0.16 \)

b). \( P(X\leq 3)=0.6 \)

c). \( P(X\leq 2)=0.214 \)