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.
Given that:
\(\lambda = 10\) which is the average of ten letters per day (the mean) and
It is a poisson distribution: \(P(X=k) = \frac {\lambda^k}{k!} e^{-\lambda}\)
\(\Rightarrow\) The probability of receiving no mail on a certain day is \(P(X=0) = \frac{10^0}{0!} e^{-10} = e^{-10}\)
The probability of receiving mail every day in the next ten years (3000 days) will be:
\(\rightarrow\) The probability of receiving some mail on a certain day is the complement of receiving no mail on a certain day: \((1-e^{-10})\)
\(\rightarrow\) The probability of receiving mail every day for 3000 days is: \((1-e^{-10})^{3000}\)
Now, the probability of having at least 1 day without any mail will be the complement of the probability of receiving mail every day in the next 3000 days: \(1-(1-e^{-10})^{3000}\)
Let’s calculate this using R-chunk:
P_oneDayNoMail <- format(1-((1-exp(-10)))^(3000), digits = 3)
#displaying the results
cat("The probability of having at least 1 day without any mail in the next 10 years is ", P_oneDayNoMail)## The probability of having at least 1 day without any mail in the next 10 years is 0.127