Question

A basketball player has made​ 70% of his foul shots during the season. If he shoots 3 foul shots in​ tonight's game, what is the probability that he makes all of the​ shots?

Answer 1

Answer:

There is a 34.3% probability that he makes all of the​ shots.

Step-by-step explanation:

For each foul shot that he takes during the game, there are only two possible outcomes. Either he makes it, or he misses. This means that we use the binomial probability distribution to solve this problem.

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:

$n = 3, p = 0.7$

What is the probability that he makes all of the​ shots?

This is P(X = 3).

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

$P(X = 3) = C_{3,3}.(0.7)^{3}.(0.3)^{0} = 0.343$

There is a 34.3% probability that he makes all of the​ shots.