Suppose each of 100 professors in a large mathematics department picks at random one of 200 courses. what is the probability that at least two professors pick the same course?

Let us say that the probability (p) of picking one of the 200 courses is:

p = 1 / 200 = 0.005

Therefore using the binomial probability (P), we can solve for the probability that at least 2 professors pick the same course. However what we will do it the opposite, calculate for the P when 0 and 1 professor only pick that course then subtract the sum from 1.

The formula for P is:

P = [n! / r! (n – r)!] p^r * q^(n – r)

where, n is the total number of professors = 100, r is the number of professors who picked the same course = 1 or 0, p = 0.005, q = 1 – p = 0.995

when r = 0

P = [100! / 0! (100 – 0)!] (0.005)^0 * (0.995)^100

P = 0.606

when r = 1

P = [100! / 1! (100 – 1)!] (0.005)^1 * (0.995)^99

P = 0.304

Therefore total P is:

P = 0.606 + 0.304

P = 0.91

Hence the probability that 2 or more will pick the same course is:

1 – 0.91 = 0.09

= 9%

Therefore there is a 0.09 or 9% probability that at least 2 will pick the same course.

step-by-step explanation: