Answer :
Answer:
0.0068% probability he will get at least 7 hits in the game
Step-by-step explanation:
For each at bat, there are only two possible outcomes. Either he gets a hit, or he does not. The probability of getting a hit in an at-bat is independent of other at-bats. So we use the binomial probability distribution 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.
A high school baseball player has a 0.159 batting average.
This means that in each at-bat, the probability of getting a hit is 0.159.
9 at bats.
This means that [tex]n = 9[/tex]
What is the probability he will get at least 7 hits in the game?
[tex]P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{9,7}.(0.159)^{7}.(0.841)^{2} = 0.000065[/tex]
[tex]P(X = 8) = C_{9,8}.(0.159)^{8}.(0.841)^{1} = 0.000003[/tex]
[tex]P(X = 9) = C_{9,9}.(0.159)^{9}.(0.841)^{0} \cong 0[/tex]
[tex]P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) = 0.000065 + 0.000003 = 0.000068[/tex]
0.0068% probability he will get at least 7 hits in the game