Question

a basketball player has a 50% chance of making each free throw. what is the probability that the player makes at least five out of six free throws?

Answer 1

To start off, we want to figure this out by getting (the amount of possibilities of five or more free throws) divided by (the amount of possibilities total). To find the total amount of possibilities, we see that there are two possibilities for each throw, making a free throw or not. Next, we have another two for the next throw, and so on, resulting in 2*2*2*2*2*2=2^6=64 total possibilities. After that, we want to figure out the chance of making at least 5 shots, meaning that the player would either have to make 5 or 6 shots. To figure that out, we can use something called combinations, which states that we can choose r amount of things from n things. For the player to make 6 shots, we have C(6,6) or combination (6,6). To expand on that, the formula is
$\frac{n!}{r!(n-r)!} = \frac{6!}{6!(6-6)!} =$
Since 6!/6!=1 and 0!=1, or answer is 1 possibility, making the possibility of getting 6/6 shots is 1/64. For 5 shots, we get C(6,5)=$\frac{6!}{5!(6-5)!} =6$ because 6!=6*5*4*3*2*1,  5!=5*4*3*2*1 , and 1!=1. Therefore, the probability of getting 5 shots out of 6 is 6/64. Adding the probabilities of getting 5 or 6 shots together, we get 7/64 as the probability of getting at least 5 shots in.

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Answer 2

For each free throw, there are only two possible outcomes, either the player makes it, or the player does not. The probability of the player making a free throw is independent of any other free throw, which means that the binomial distribution is used to solve this question.

Doing this, we get that: 0.1094 = 10.94% probability that the player makes at least five out of six free throws.

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.

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A basketball player has a 50% chance of making each free throw

This means that $p = 0.5$

6 free throws:

This means that $n = 6$

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What is the probability that the player makes at least five out of six free throws?

This is:

$P(X \geq 5) = P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{6,5}.(0.5)^{5}.(0.5)^{1} = 0.0938$

$P(X = 6) = C_{6,6}.(0.5)^{6}.(0.5)^{0} = 0.0156$

Then

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0938 + 0.0156 = 0.1094$

Thus, 0.1094 = 10.94% probability that the player makes at least five out of six free throws.

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