Answer:
The probability is 0.1381, or about 13.8%
Step-by-step explanation:
The probability of a count k out of n with a known event probability p is described by the binomial distribution:
[tex]p(k) = {{n}\choose{k}}p^k\cdot (1-p)^{n-k}[/tex]
We are given n=100, p=0.91 and are asking for the probability of seeing exactly k=91 positive events (enrollment), so:
[tex]p(91) = {{100}\choose{91}}0.91^{91}\cdot0.09^{9} = 0.1381[/tex]
