MATH SOLVE

4 months ago

Q:
# 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?

Accepted Solution

A:

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

[tex] \frac{n!}{r!(n-r)!} = \frac{6!}{6!(6-6)!} =[/tex]

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)=[tex] \frac{6!}{5!(6-5)!} =6[/tex] 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.

Feel free to ask further questions!

[tex] \frac{n!}{r!(n-r)!} = \frac{6!}{6!(6-6)!} =[/tex]

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)=[tex] \frac{6!}{5!(6-5)!} =6[/tex] 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.

Feel free to ask further questions!