What is the probability that a 93% FT shooter will make 9 FT out of 10 FTA?
Confused ..
Given probability of success on any one free throw is p=0.93, the probability of geting exacty k free throws in n attemps is:
$\displaystyle P(\text{k\ success\ in \ n \ attempts}) = C(n,k)p^k(1-p)^{n-k}$
where C(n,k) is the number of combinations of k items in a set of n:
$\displaystyle C(n,k) = \frac {n!}{k!(n-k)!}$
So here you have:
$\displaystyle P(\text{9\ success\ in\ 10\ attempts}) = C(10,9)(0.93)^9(0.07)^1 = 0.3643$