So, we're looking for $\displaystyle P(X\geq 2)$ ...Office gets each day new mail according to Poisson distribution $\displaystyle P(\lambda)$.

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?

$\displaystyle P(X\geq 2)=1-P(X=0)-P(X=1)=1-...$

P(X=0) - 5 ways to be left with0mail atday 3(# income mail onday 1, # income mail onday 2): (0,0) (0,1) (1,0) (1,1) (2,0)

P(X=1) - 3 ways to be left with1mail atday 3: (0,2) (1,2) (2,1)

Now, how do you use above with Poisson formula?

Official solution is $\displaystyle 1-e^{-2\lambda}(1+2\lambda+2\lambda^2+\frac{7\lambda^3}{ 6})$.