Question

A high school baseball player has a 0.212 batting average. In one game, he gets 9 at bats. What is the probability he will get at least 2 hits in the game?

Answer 1

59.92% probability he will get at least 2 hits in the game.

Explanation:

For each at bat, there are only two possible outcomes. Either he gets a hit, or he does not. The probability of a getting a hit in each at bat is independent. 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this problem we have that:

`p = 0.212`

In one game, he gets 9 at bats. What is the probability he will get at least 2 hits in the game?

This is
`P(X \geq 2)` when
`n = 9`

He either gets less than two hits in the game, or he gets at least two hits. The sum of the probabilities of these events is decimal 1. So

`P(X < 2) + P(X \geq 2) = 1`

`P(X \geq 2) = 1 - P(X < 2)`

In which

`P(X < 2) = P(X = 0) + P(X = 1)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(9,0).(0.212)^(0).(0.788)^(9) = 0.1171`

`P(X = 1) = C_(9,1).(0.212)^(1).(0.788)^(8) = 0.2837`

`P(X < 2) = P(X = 0) + P(X = 1) = 0.1171 + 0.2837 = 0.4008`

Finally

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4008 = 0.5992`

59.92% probability he will get at least 2 hits in the game.