Question

In an NBA championship series, the team that wins four games out of seven is the winner. Suppose that teams ???? and ???? face each other in the championship games and that team ???? has a probability 0.55 of winning a game over team B. a. What is the probability that team ???? will win the series in 6 games? b. What is the probability that team ???? will win the series? c. If teams ???? and ???? were facing each other in a regional playoff series, which is decided by winning three out of 5 games, what is the probability that team ???? would win the series?

Answer 1

Answer:

a) P=0.185

b) P=0.608

c) P=0.593

Step-by-step explanation:

The question is not complete

"In an NBA 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 a probability 0.55 of winning a game over team B.

a. What is the probability that team A will win the series in 6 games?

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

c. If teams A and B were facing each other in a regional playoff series, which is decided by winning three out of 5 games, what is the probability that team A would win the series?

a) It is implicit that 6 games are played. The number of games needed to win a series in 6 games is 4, and the last winning should be in the 6th game.

So we have 5 random results in the first 5 games in which A team wins 3 matches and losses 2 matches.

This can be modeled as a binomial distribution with k=3, n=5 and p=0.55.

$P(k=3)=\frac{n!}{k!(n-k)!}p^k*(1-p)^{n-k}\\\\P(k=3)=\frac{5!}{3!2!}*0.55^3*0.45^2=10*0.166*0.203=0.337$

Then, this probability has to be multiplied by the probability of winning the last game.

$P(W_6)=P(k=3;n=5)*P(W_{6th})=0.337*0.55=0.185$

The probability that team A will win the series in 6 games is P=0.185.

b) In this case, we have to calculate the probability of A winning at least 4 games. When a team reaches 4 wins, the series is over. This can happen in 4, 5, 6 or 7 games.

The probability of winning the series is:

$P(W)=[P(3;3)+P(3;4)+P(3;5)+P(3;6)]*P(win\,last\,game)$

Probability of 3 wins in 3 games

$P(3;3)=\frac{3!}{3!0!}*0.55^3*0.45^0 =1*0.166*1=0.166$

Probability of 3 wins in 4 games

$P(3;4)=\frac{4!}{3!1!}*0.55^3*0.45^1 =4*0.166*0.45=0.299$

Probability of 3 wins in 5 games

$P(3;5)=\frac{5!}{3!2!}*0.55^3*0.45^2 =10*0.166*0.203=0.337$

Probability of 3 wins in 6 games

$P(3;6)=\frac{6!}{3!3!}*0.55^3*0.45^3 =20*0.166*0.091=0.303$

Then

$P(W)=[P(3;3)+P(3;4)+P(3;5)+P(3;6)]*P(win\,last\,game)\\\\P(W)=[0.166+0.299+0.337+0.303]*0.55=1.105*0.55=0.608$

The probability of team A of winning the series is P=0.608.

c) The probabilty of winning the regional playoff is

$P(W)=[P(2;2)+P(2;3)+P(2;4)]*P(win\,last\,game)$

$P(2;2)=\frac{2!}{2!0!}*0.55^2*0.45^0 =1*0.303*1=0.303\\\\P(2;3)=\frac{3!}{2!1!}*0.55^2*0.45^1 =3*0.303*0.45=0.408\\\\P(2;4)=\frac{4!}{2!2!}*0.55^2*0.45^2 =6*0.303*0.203=0.368$

Then,

$P(W)=[P(2;2)+P(2;3)+P(2;4)]*P(win\,last\,game)\\\\P(W)=[0.303+0.408+0.368]*0.55=1.079*0.55=0.593$