Respuesta :
Answer:
a)
There is a 12.96% probability that the Cubs win the World Series in exactly four games.
There is a 20.736% probability that the Cubs win in 5 games.
There is a 20.736% probability that the Cubs win in 6 games.
There is a 16.60% probability that the Cubs win in 7 games.
b)
There is a 64.8% probability that the Cubs would win in a 2-out-of-3 series.
Step-by-step explanation:
For each game, there are these following probabilities:
A 60% probability that the Cubs win.
A 40% probability that the Blue Jays win.
The combination formula is important to solve this problem:
[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
a.Find the probability, for i = 4, 5, 6, 7, That the cubs win the World Series in exactly i games.
i = 4.
This is the probability that the Cubs win in 4 games. So
[tex]P = (0.6)^{4} = 0.1296[/tex]
There is a 12.96% probability that the Cubs win the World Series in exactly four games.
i = 5.
This is the Cubs winning four games and the Blue Jays 1. The Blue Jays win cannot happen in the fifth game, so the number of possibilities is a combination of 4 by 3. So
[tex]C_{4,,3} = \frac{4!}{3!(4-3)!} = 4[/tex]
The probability that the Cubs win in 5 games is:
[tex]P = 4*(0.6)^4*(0.4) = 0.20736[/tex]
There is a 20.736% probability that the Cubs win in 5 games.
i = 6.
This is the Cubs winning four games and the Blue Jays 2. The Blue Jays cannot win game 6, so the number of possibilities is a combination of 5 by 3.
[tex]C_{5,3} = \frac{5!}{3!(5-3)!} = 10[/tex]
The probability that the Cubs win in 6 games is:
[tex]P = 10*(0.6)^4*(0.4)^2 = 0.20736[/tex]
There is a 20.736% probability that the Cubs win in 6 games.
i = 7.
This is the Cubs winning four games and the Blue Jays 3. The Blue Jays cannot win game 7, so the number of possibilities is a combinaiton of 6 by 3.
[tex]C_{6,3} = \frac{6!}{3!(5-3)!} = 20[/tex]
The probability that the Cubs win in 7 games is:
[tex]P = 20*(0.6)^4*(0.4)^3 = 0.1660[/tex]
There is a 16.60% probability that the Cubs win in 7 games.
b.What is the probability that the Cubs would win in a 2-out-of-3 series instead ?
This is the sum of the probabilities that they win in 2 games and the they win in 3 games.
i = 2.
This is the Cubs winning both games. So
[tex]P_{1} = 0.6*0.6 = 0.36[/tex]
i = 3.
This is the Cubs winning 2 games and the Blue Jays 1. The Blue Jays cannot win game 3. So the number of possibilities is a combination of 2 by 1.
[tex]C_{2,1} = \frac{2!}{1!(2-1)!} = 2[/tex]
And the probability is:
[tex]P_{2} = 2*0.6*0.6*0.4 = 0.288[/tex]
Finally
[tex]P = P_{1} + P_{2} = 0.36 + 0.288 = 0.648[/tex]
There is a 64.8% probability that the Cubs would win in a 2-out-of-3 series.
