Question

a local community college determines the probability that a student will reenroll for a second year is 0.91. A representative surbeys 100 random first year students and asks them if they will be enrolling for the next year. What is the probability that exactly 91 of them wil enroll?

Answer 1

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:

$p(k) = {{n}\choose{k}}p^k\cdot (1-p)^{n-k}$

We are given n=100, p=0.91 and are asking for the probability of seeing exactly k=91 positive events (enrollment), so:

$p(91) = {{100}\choose{91}}0.91^{91}\cdot0.09^{9} = 0.1381$