According to the Canadian Journal of Information and Library

According to the Canadian Journal of Information and Library Science (Vol 33, 2009), the probability that workers in Canadian law libraries are satisfied with their job is 0.9. In a random sample of 20 law librarians in Canada, what is the probability that exactly 17 of them are satisfied with their job?

Solution

Given X follows Binomial distribution with n=20 and p=0.9

P(X=x)=20Cx*(0.9^x)*(0.1^(20-x)) for x=0,1,2,..,20

So the probability is

P(X=17) = 20C17*(0.9^17)*(0.1^(20-17)) =0.1901199