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:

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

Step-by-step 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.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

Makes 56% of free throws.

So

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

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

just needed some points but try socratic

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step-by-step explanation: