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

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2 years ago