Probability of $\text{Pr}(X+Y=k)$ for two independent random variables $X$ and $Y$ Let $X$ and $Y$ denote two independent random variables with $$\text{Pr}(X=k)=\begin{pmatrix}m\\k\end{pmatrix}p^k(1-p)^{m-k}\;\;\;\text{and}\;\;\;\text{Pr}(Y=k)=\begin{pmatrix}n\\k\end{pmatrix}p^k(1-p)^{n-k}$$
I want to show, that it holds $$\text{Pr}(X+Y=k)=\begin{pmatrix}m+n\\k\end{pmatrix}p^k(1-p)^{n+m-k}$$ for all $1\le k\le m+n$.
Proof:
\begin{equation}
\begin{split}
\text{Pr}(X+Y=k)&=\sum_{l=0}^m\text{Pr}(X=l)\text{Pr}(X+Y=k\;|\;X=l)\\
&=\sum_{l=0}^m\text{Pr}(X=l)\text{Pr}(Y=k-l)\\
&=p^k(1-p)^{m+n-k}\sum_{l=0}^m\begin{pmatrix}m\\l\end{pmatrix}\begin{pmatrix}n\\k-l\end{pmatrix}
\end{split}
\end{equation}
While this looks good, I'm unable to proceed from here.
 A: From the equation in the post as revised, all we need to show is that
$$\sum_{l=0}^m \binom{m}{l}\binom{n}{k-l}=\binom{m+n}{k}.$$
This is easiest done combinatorially. We have $m$ boys and $n$ girls in a group. We want to choose $k$ people. 
The number of ways to do this is clearly $\binom{m+n}{k}$. Let us count another way.
There are $\binom{m}{l}$ to choose $l$ boys, and for every such way there are $\binom{n}{k-l}$ ways to choose $k-l$ girls to bring the total up to $k$. 
The number of ways to choose $l$ boys and $k-l$ girls is therefore $\binom{m}{l}\binom{n}{k-l}$. Add up, $l=0$ to $l=m$. We get that the number of ways to choose $k$ people is $\sum_{l=0}^m \binom{m}{l}\binom{n}{k-l}=\binom{m+n}{k}$.
We have counted the number of ways to choose $k$ people in two different ways. Since the answers must be the same, we have proved the required identity. 
Remark: For some values of $m,n,k,l$, we may have $k-l\gt n$ or $k-l\lt 0$.  The formula still works, if we define $\binom{a}{b}$ to be $0$ if $a\lt b$ or if $b$ is negative. 
A: Both of your rvs are independent Binomial ($n,p$). Look at the probabilities and the number of ways these rvs take on the same value, e.g. 
$$
0: P(X=0 \cap Y=0) = P(X=0) \cdot P(Y=0)\\
1: P(X=0 \cap Y=1) + P(X=1 \cap Y=0) = P(X=1) \cdot P(Y=0) + P(X=0) \cdot P(Y=1)
$$
and so on, you'll see the pattern soon. This is called convolution. The result will be Binomial distribution $(2n, p$).
