The probability of 0 potholes is e^{-6}*6^{0}/0! = e^{-6}.

The probability of 1 pothole is e^{-6}*6^{1}/1! = 6*e^{-6}.

The probability of 2 potholes is e^{-6}*6^{2}/2! = 18*e^{-6}.

The probability of 3 potholes is e^{-6}*6^{3}/3! = 36*e^{-6}.

Take the sum the answer is 61*e^{-6} =~ 0.151204 .

This problem was taken from the May 1982 Actuary Exam 110, problem 20.

Michael Shackleford, ASA, August 17, 1999