Question

Suppose a basketball player typically makes five out of eight foul shots.

Question : How could you predict the probability of the player making at least one shot out of 3 foul shot attempts?

Answer 1

0.9473 is the probability that the player making at least one shot out of 3 foul shot attempts.

Explanation:

We are given the following information:

We treat basketball player making a foul shot as a success.

P(Foul shot) =

`(5)/(8) = 0.625`

Then the number of adults follows a binomial distribution, where

`P(X=x) = \binom{n}{x}.p^x.(1-p)^(n-x)`

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 3

P(at least one shot out of 3 foul shot attempts)

We have to evaluate:

`P(x \geq 1) =1- P(x = 0)\\\\=1- \binom{3}{0}(0.625)^0(1-0.625)^3\\\\= 1 - 0.0527\\= 0.9473`

0.9473 is the probability that the player making at least one shot out of 3 foul shot attempts.