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

Answer:

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

Answer:

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

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.