= Solution
For <independent random variables>, their sum has the product of their <probability generating functions>:
$$
G_{X+Y}(z)=\mathbb E[z^Xz^Y]=G_X(z)G_Y(z)=(1-p+pz)^{n+m}.
$$
Comparing polynomial coefficients identifies a <binomial distribution> with parameters $n+m,p$:
$$
\boxed{\mathbb P(X+Y=k)=\binom{n+m}{k}p^k(1-p)^{n+m-k},\qquad k=0,\ldots,n+m,}
$$
and the <probability> is zero at all other values. The common success <probability> is essential to this binomial form; independent binomial counts with different <probabilities> do not in general add to a binomial count.
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