The probability that Shruti succeeds at any given free-throw is 80%, percent. She was curious how many free-throws she can expect to succeed in a sample of 12 free-throws.

She simulated 25 samples of 12 free-throws where each free-throw had a 0.8, point, 8 probability of being a success.

Shruti counted how many free-throws were successes in each simulated sample. Here are her results:

Use her results to estimate the probability that she succeeds at 10 or more free-throws in a sample of 12 free-throws.

Give your answer as either a fraction or a decimal.

0.5584 probability that she succeeds at 10 or more free-throws in a sample of 12 free-throws.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either she makes it, or she 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.

The probability that Shruti succeeds at any given free-throw is 80%, percent.

This means that

Sample of 12 free throws:

This means that

Use her results to estimate the probability that she succeeds at 10 or more free-throws in a sample of 12 free-throws.

In which

0.5584 probability that she succeeds at 10 or more free-throws in a sample of 12 free-throws.

a = 11 : 4

b = 38

c = 1 for every 6