Question

Suppose that each time Giannis Antetokounmpo shoots a free throw, he has a 3/4 probability of success. If Giannis shoots three free throws, what is the probability that he succeeds on at least two of them

Answer 1

Answer:

84.38% probability that he succeeds on at least two of them

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either Giannis makes it, or he does not. The free throws are 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.

He has a 3/4 probability of success.

This means that $p = \frac{3}{4} = 0.75$

Giannis shoots three free throws

This means that $n = 3$

What is the probability that he succeeds on at least two of them

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

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

$P(X = 2) = C_{3,2}.(0.75)^{2}.(0.25)^{1} = 0.4219$

$P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.4219$

$P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4219 + 0.4219 = 0.8438$

84.38% probability that he succeeds on at least two of them