Since $X$ and $Y$ are independent $\operatorname{Bin}(n,p)$ , their sum is $X+Y\sim\operatorname{Bin}(2n,p)$ . Using independence in the numerator,

$\mathbb{P}(X=j\mid X+Y=k) =\frac{\mathbb{P}(X=j)\,\mathbb{P}(Y=k-j)}{\mathbb{P}(X+Y=k)} =\frac{\binom{n}{j}p^{j}q^{n-j}\;\binom{n}{k-j}p^{k-j}q^{n-k+j}} {\binom{2n}{k}p^{k}q^{2n-k}} .$

The powers of $p$ combine to $p^{k}$ and those of $q$ to $q^{2n-k}$ , so everything depending on $p$ cancels:

$\mathbb{P}(X=j\mid X+Y=k)=\frac{\binom{n}{j}\binom{n}{k-j}}{\binom{2n}{k}}, \qquad \max(0,k-n)\le j\le\min(n,k).$

This is the hypergeometric distribution — the conditional law is free of $p$ , which is exactly why $X+Y$ is sufficient for $p$ . Option (D) is the only choice with this structure, and is the intended key.

Note on a misprint. As printed, option (D) has $\binom{2n}{n}$ in the denominator. The correct normalising constant is $\binom{2n}{k}$ ; the two agree only when $k=n$ . Vandermonde's identity $\sum_{j}\binom{n}{j}\binom{n}{k-j}=\binom{2n}{k}$ confirms that $\binom{2n}{k}$ is what makes the probabilities sum to one.