Proving Binomial Random Variable Identity Good Morning All, 
May I ask for a clue to the following problem? 
I got stuck and I am now wondering if I understood the problem correctly.

Let X, Y be independent random variables bin(m, p), bin(n, p) respectively. 
Show that X+Y is a binomial random variable with parameter (m+n, p)

So $\mathbb P (X = i) = {m \choose i}{p^i}{q^{m-i}}$, $\mathbb P (Y = j) = {n \choose j}{p^j}{q^{n-j}}$
Let $m \le n$, $\mathbb P (X+Y = t) = $ 
$ \mathbb P (X=0, Y=t) + \mathbb P (X=1, Y=t-1) ... + \mathbb P (X=m, Y=t-m) $
$ = \sum_{i=1}^m \mathbb P (X=i) \mathbb P (Y=t-i)$ 
$ = \sum_{1=1}^m {m \choose i}{n \choose t-i} p^{i+t-i}q^{m+n-i-t+i}$

Now, I am stuck with the combinations.  I can't get them to look like $m+n \choose t$.
 A: I'm sure by some books this won't be accepted as a full proof, and it doesn't answer your question, but here's a less formal proof:
Say you have $m$ identical experiments in one room, and $n$ of the same experiment in another.  Also, assume that they are all independent, and that each of them have probability of success $p$. Then $X$ is the number of successful experiments in the first room and $Y$ the number of successful experiments in the other. $X + Y$ will then intuitively be the number of successful experiments in the two rooms together. This is then $m+n$ identical, independent experiments, each with success probability $p$, and it is therefore bin$(m+n, p)$.
A: You should in your third line only go to $X=t$, $Y=0$. So you want to show that
$$\sum_{i=0}^t \binom{m}{i}\binom{n}{t-i}=\binom{m+n}{t}.$$
Here is combinatorial proof of the above identity. We have a group of $m$ boys and $n$ girls, and want to pick $t$ people. 
By definition, this can be done in $\dbinom{m+n}{t}$ ways.
Let us count this another way. We can pick $0$ boys and $t$ girls. This can be done in $\binom{m}{0}\binom{n}{t}$ ways.
Or else we can pick $1$ boy and $t-1$ girls. This can be done in $\binom{m}{1}\binom{n}{t-1}$ ways. 
Or else we can pick $2$ boys and  $t-2$ girls. This can be done in $\binom{m}{2}\binom{n}{t-2}$ ways.
Continue, and add up.  We get the desired identity.
Note: If $m$ is "small" and $t$ and $n$ are largish, some of the $i$ in the sum may be greater than $m$. The above expression is still correct, if we use the convention that $\binom{a}{b}=0$ if $a$ and $b$ are non-negative integers such that $a\lt b$. 
