So, we're looking for ...Office gets each day new mail according to Poisson distribution .
Days are independent.
Each day, one mail gets answered (if there is any), other mails (if there are any) are left for later days.
What is the probability, that after 2 days, the office is left with at least 2 un-answered mails?
P(X=0) - 5 ways to be left with 0 mail at day 3 (# income mail on day 1, # income mail on day 2): (0,0) (0,1) (1,0) (1,1) (2,0)
P(X=1) - 3 ways to be left with 1 mail at day 3: (0,2) (1,2) (2,1)
Now, how do you use above with Poisson formula?
Official solution is .