Question

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.

Answer 1

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

hope this helps!! <3

Answer 2

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

Explanation:

For each free throw, there are only two possible outcomes. Either it is a success, or it is not. The probability of a free throw being a success is independent of other free throws. So we use the binomial probability distibution 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.

Sample of 12 free throws.

This means that
`n = 12`

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

This means that
`p = 0.8`

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

`P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)`

Then

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

`P(X = 10) = C_(12,10).(0.8)^(10).(0.2)^(2) = 0.2835`

`P(X = 11) = C_(12,11).(0.8)^(11).(0.2)^(1) = 0.2062`

`P(X = 12) = C_(12,12).(0.8)^(12).(0.2)^(0) = 0.0687`

`P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) = 0.2835 + 0.2062 + 0.0687 = 0.5584`

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