Find $E(X+Y)$ when $X$ and $Y$ have a mean from 1 to 100 
There are 100 balls in a bag, each numbered 1 through 100 (uniquely). Two of them are drawn at random. We then choose a Poisson random variable $X$ with mean equal to the number on the first ball that was drawn and a Poisson random variable $Y$ with the mean equal to the number on the second ball that was drawn. State the expected value of $X + Y$.

I think that $E(X+Y) = 101$ but I'm not sure. Can anyone explain why this is the answer if it is indeed correct?
 A: Hint 1: $E[X+Y]=E[X] + E[Y]$. So it suffices to compute each expectation individually.
Hint 2: What is $E[X \mid \text{first ball is $\lambda$}]$? What is $P(\text{first ball is $\lambda$})$? The apply the law of total expectation to compute $E[X]$.
A: $\newcommand{\E}{\mathbb{E}} \newcommand{\P}{\mathbb{P}}$
An explanation. Let $M$ be the number drawn first and $N$ be the number drawn second. You have $M\sim \text{Unif}\{1,2,...,100\}$ and $N|M=m\sim\text{Unif}\{1,2,...,100\}\setminus \{m\}$. And you have $X\sim \text{Pois}(M)$ and $Y\sim\text{Pois}(N)$. Now we try to find the distribution of $N$:
\begin{align}
\P(N=n)=\sum_{m=1, m\neq n}^{100}\P(N=n|M=m)\P(M=m) = \sum_{m=1, m\neq n}^{100} \frac{1}{99}\frac{1}{100} = \frac{1}{100}
\end{align}
Hence $N\sim\text{Unif}\{1,2,..,100\}$.
We have $\E[X]=\E[\E[X|M]]$ and $\E[Y]=\E[\E[Y|N]]$
Furthermore 
\begin{align}
\E[X|M]=M ,\ \ \ \ \E[Y|N]=N
\end{align}
So:
\begin{align}
\E[X] = \E[M]=\frac{100+1}{2} =\frac{101}{2}
\end{align}
and
\begin{align}
\E[Y] = \E[N]=\frac{100+1}{2} =\frac{101}{2}
\end{align}
And $\E[X+Y]=\E[X]+\E[Y]=101$ as expected. 
