probability problem with Poisson distribution Problem
A retailer knows that the demand of boxes is a random variable with Poisson distribution of parameter $\lambda=2$ boxes per week. The retailer completes his stock on monday so as to have four boxes at the beginning of the week. 
a) What is the minimum number of boxes he should have at the beginning of each week so as the probability to satisfy all the demand of boxes is greater than or equal to $0.99$?
b) What is the distribution of the number of sold boxes per week?
My attempt at a solution
a) If $X$ is the number of boxes demanded per week, then $X \sim P(2)$, let $n$ be the stock at the beginning of a week, in order to be able to satisfy all the orders of boxes of the week $X$ has to be less than or equal to $n$. So, we want to find $n$ such that $$P(X \leq n) \geq 0.99$$
But $$P(X \leq n)=\sum_{i=0}^n \dfrac{e^22^i}{i!}$$
I have no idea how to find the minimum value of $n$ from the inequality $$\sum_{i=0}^n \dfrac{e^22^i}{i!} \geq 0.99$$
As for part b), if $Y=\text{number of sold boxes per week}$, then $Y=X$ for $X \leq 4$ and $Y=4$ for $X >4$, I don't know how to figure out the distribution of $Y$ from this information.
Any help would be appreciated. Thanks in advance.

You can solve it by inspection.   Just fill in the table:
$$\begin{array}{l:l}x & \dfrac{2^x~\mathsf e^{-2}}{x!} & \displaystyle\sum\limits_{k=1}^x\dfrac{2^k~\mathsf e^{-2}}{k!}
\\ \hdashline 0 \\ 1 \\ 2 \\ 3 \\ 4\\ 5 \\ \hline 6 \\ \vdots
\end{array}$$
Stop when the third column exceeds $0.99$.
  Do you have access to a spreadsheet program?

for b) since you can't sell more boxes than stocked, if the answer to (a) is $n$ then
$$\mathsf P(Y=y\mid X=n) ~=~ \begin{cases}{2^y\mathsf e^{-2}}/{y!} & : y\in\{0,\ldots, n-1\} \\[1ex] 1-\sum_{k=0}^n ({2^k~\mathsf e^{-2}}/{k!}) & : y=n\\[1ex] 0 & : \textsf{other }y\end{cases}$$
Where we know $\mathsf P(Y=n\mid X=n) \leq 0.01$ for the required $n$.
