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

Answer:

59.92% probability he will get at least 2 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 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.

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

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.