Assume that a procedure yields a binomial distribution with

Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. Round to three decimal places.
n = 15, x = 11, p = 0.85

Solution

We want the expansion of (p+q)^5 for the term containing p^2

This term is in the form of B(N,K) p^2q^3

B(N,K) = N! / K!(N-K)!
B(15,11) = 15! / 11!4! = 1365

So the value of the term is 1365 * 0.85^11 * 0.15^4 = 0.115