Answer: Our required probability is 0.175.
Step-by-step explanation:
Since we have given that
Average of 5 e-mails messages per day.
so, [tex]\lambda=5[/tex]
We need to find the probability that on a randomly selected day your professor will have 5 messages.
Let X be the number of messages in poisson distribution.
So, P(X=5) is given by
[tex]\dfrac{\lambda^xe^{-\lambda}}{x!}\\\\=\dfrac{5^5e^{-5}}{5!}\\\\=0.175[/tex]
Hence, our required probability is 0.175.