Question

A high school baseball player has a 0.159 batting average. In one game, he gets 9 at bats. What is the probability he will get at least 7 hits in the game?

Answer 1

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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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 $n = 9$

What is the probability he will get at least 7 hits in the game?

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{9,7}.(0.159)^{7}.(0.841)^{2} = 0.000065$

$P(X = 8) = C_{9,8}.(0.159)^{8}.(0.841)^{1} = 0.000003$

$P(X = 9) = C_{9,9}.(0.159)^{9}.(0.841)^{0} \cong 0$

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) = 0.000065 + 0.000003 = 0.000068$

0.0068% probability he will get at least 7 hits in the game