Respuesta :
Answer:
a) 0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.
b) The expected number of at-bats until the next home run is 6.25.
Step-by-step explanation:
For each at bat, there are two possible outcomes. Either it is a home run, or it is not. The probability of an at bat resulting in a home run is independent of any other at-bat, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The probability that Barry Bonds hits a home run on any given at-bat is 0.16
This means that [tex]p = 0.16[/tex]
Part A: What is the probability that the next home run will be on his fifth at-bat?
0 on his next 4(P(X = 0) when n = 4)
Home run on his 5th at-bat, with 0.16 probability. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{4,0}.(0.16)^{0}.(0.84)^{4} = 0.49787136 [/tex]
0.49787136 *0.16 = 0.0797.
0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.
Part B: What is the expected number of at-bats until the next home run?
The expected number of trials for n successes is given by:
[tex]E = \frac{n}{p}[/tex]
In this question, [tex]n = 1, p = 0.16[/tex]. So
[tex]E = \frac{1}{0.16} = 6.25[/tex]
The expected number of at-bats until the next home run is 6.25.
