Let $X$ and $Y$ be independent RV's such that $X\sim\text{Binomial}(m,p)$ and $Y\sim\text{Binomial}(n,p)$. Determine
(a) $\textbf{P}(X = x \mid X+Y = s)$
(b) $\textbf{E}(X\mid X + Y)$ and $\textbf{V}(X \mid X+Y)$
(c) Check that $\textbf{E}\{\textbf{E}(X\mid X+Y)\} = \textbf{E}(X)$
MY ATTEMPT
Firsly, I would like to determine the distribution of $Z = X+Y$. Since $X$ and $Y$ are independent, \begin{align*} \textbf{P}(Z = z) = \textbf{P}(X + Y = z) = \sum_{x=0}^{z}\textbf{P}(X = x,Y = z - x) = \sum_{x=0}^{z}\textbf{P}(X = x)\textbf{P}(Y = z-x) \end{align*}
Consequently, we have
\begin{align*} p_{Z}(z) & = \textbf{P}(Z = z) = \sum_{x=0}^{z}{n\choose z-x}p^{z-x}(1-p)^{n-z+x}{m\choose x}p^{x}(1-p)^{m-x}\\ & = \sum_{x=0}^{z}{n\choose z-x}{m\choose x}p^{z}(1-p)^{m + n - z} = {m+n\choose z}p^{z}(1-p)^{m+n-z} \end{align*}
Another possible approach is to consider each Binomial distribution as a sum of Bernoulli distributions. Therefore $Z\sim\text{Binomial}(m+n,p)$. Based on this, could someone help me out?
EDIT
(a) As a consequence from the independence between $X$ and $Y$, we have \begin{align*} \textbf{P}(X = x | X + Y = s) & = \frac{\textbf{P}(X = x, X+Y = s)}{\textbf{P}(X+Y = s)} = \frac{\textbf{P}(X = x)\textbf{P}(Y = s - x)}{\textbf{P}(X+Y = s)}\\\\ & = \frac{\displaystyle{m\choose x}{n\choose s-x}p^{s}(1-p)^{m+n-s}}{\displaystyle{m+n\choose s}p^{s}(1-p)^{m+n-s}} = \frac{\displaystyle{m\choose x}{n\choose s - x}}{\displaystyle{m+n\choose s}} \end{align*}
(b) According to wikipedia (Hypergeometric random variable), we have
\begin{align*} \textbf{E}(X | X + Y) = \frac{sm}{m+n}\quad\text{and}\quad\textbf{V}(X | X+Y) = \ldots \end{align*}
(c) Finally, we have
\begin{align*} \textbf{E}\{\textbf{E}(X|X+Y)\} = \frac{m}{m+n}\textbf{E}(X+Y) = \frac{m(m+n)p}{m+n} = mp = \textbf{E}(X) \end{align*}