Question

Steph makes 90 % 90%90, percent of the free throws she attempts. She is going to shoot 3 33 free throws. Assume that the results of free throws are independent from each other. Let X XX represent the number of free throws she makes. Find the probability that Steph makes exactly 1 11 of the 3 33 free throws. You may round your answer to the nearest hundredth. P ( X = 1 ) =

Answer 1

Using the binomial distribution, it is found that there is a 0.027 = 2.7% probability that he makes exactly 1 of the 3 free throws.

For each free throw, there are only two possible outcomes, either he makes it, or he misses it. The results of free throws are independent from each other, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• He makes 90% of the free throws, hence $p = 0.9$.

• He is going to shoot 3 free throws, hence $n = 3$.

The probability that he makes exactly 1 is P(X = 1), hence:

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

$P(X = 1) = C_{3,1}.(0.9)^{1}.(0.1)^{2} = 0.027$

0.027 = 2.7% probability that he makes exactly 1 of the 3 free throws.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377