Respuesta :
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.
[tex]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[/tex]
Then, this probability has to be multiplied by the probability of winning the last game.
[tex]P(W_6)=P(k=3;n=5)*P(W_{6th})=0.337*0.55=0.185[/tex]
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:
[tex]P(W)=[P(3;3)+P(3;4)+P(3;5)+P(3;6)]*P(win\,last\,game)[/tex]
Probability of 3 wins in 3 games
[tex]P(3;3)=\frac{3!}{3!0!}*0.55^3*0.45^0 =1*0.166*1=0.166[/tex]
Probability of 3 wins in 4 games
[tex]P(3;4)=\frac{4!}{3!1!}*0.55^3*0.45^1 =4*0.166*0.45=0.299[/tex]
Probability of 3 wins in 5 games
[tex]P(3;5)=\frac{5!}{3!2!}*0.55^3*0.45^2 =10*0.166*0.203=0.337[/tex]
Probability of 3 wins in 6 games
[tex]P(3;6)=\frac{6!}{3!3!}*0.55^3*0.45^3 =20*0.166*0.091=0.303[/tex]
Then
[tex]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[/tex]
The probability of team A of winning the series is P=0.608.
c) The probabilty of winning the regional playoff is
[tex]P(W)=[P(2;2)+P(2;3)+P(2;4)]*P(win\,last\,game)[/tex]
[tex]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[/tex]
Then,
[tex]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[/tex]
