Question

A college basketball player makes 5 6 56 of his free throws. Assuming free throws are independent, the probability that he makes exactly three of his next four free throws is:

Answer 1

30.91% probability that he makes exactly three of his next four free throws

Explanation:

For each free throw, there are only two possible outcomes. Either he makes ir, or he does not. The probability of making a free throw is independent of other free throws. 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Makes 56% of free throws.

So
`p = 0.56`

Assuming free throws are independent, the probability that he makes exactly three of his next four free throws is:

This is P(X = 3) when n = 4. So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 3) = C_(4,3).(0.56)^(3).(0.44)^(1) = 0.3091`

30.91% probability that he makes exactly three of his next four free throws