When John Kemeny was chair of the Mathematics Department at Dartmouth College, he received an average of ten letters each day. On a certain weekday he received no mail and wondered if it was a holiday. To decide this he computed the probability that, in ten years, he would have at least 1 day without any mail. He assumed that the number of letters he received on a given day has a Poisson distribution. What probability did he find? Hint: Apply the Poisson distribution twice. First, to find the probability that, in 3000 days, he will have at least 1 day without mail, assuming each year has about 300 days on which mail is delivered.

The probability of no mail in a day when µ=10 and x=0 is P(10,0)=(e^-10)(10^0)/0!=e^-10=0.0000454 approx. This is 1/22026 approx., or an expectation of one day without mail in about 22,026 days. If we scale this up for 3000 days we get 3000*0.0000454=0.1362. So a new Poisson distribution can be set up where µ=0.1362. P(0.1362,x)=(e^-0.0000454)0.0000454^x/x! where x is 0, 1, 2, days, etc.

Also, if x=0, (no days without mail) then the probability of at least one day in 3,000 days without mail is 1-P(0.1362,0)=1-e^-0.1362=1-0.8726=0.1273 or 12.73% approx.

by Top Rated User (616k points)