Question

A basketball player steps to the line to shoot three free-throws. If her free-throw completion average is .92, and we assume her probability of completion is constant and each shot is independent, what is the probability that she completes at least two free-throws? (Round your answer to three decimal places.)

Answer 1

Answer:

There is a 98.2% probability that she completes at least two free-throws.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either she makes it, or she misses. This means that we can solve this problem using concepts of the binomial probability distribution.

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 combinatios 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:

There are three free-throws, so $n = 3$.

Her completion average is 0.92, so $p = 0.92$

What is the probability that she completes at least two free-throws?

This is

$P(X \geq 2) = P(X = 2) + P(X = 3)$

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

$P(X = 2) = C_{3,2}.(0.92)^{2}.(0.08)^{1} = 0.203$

$P(X = 3) = C_{3,3}.(0.92)^{3}.(0.08)^{0} = 0.779$

$P(X \geq 2) = P(X = 2) + P(X = 3) = 0.203 + 0.779 = 0.982$

There is a 98.2% probability that she completes at least two free-throws.