Question

In: Statistics and Probability

Discrete R.V and Probability Distribution

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?

Solutions

Expert Solution

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

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