Binomial-binomial is binomial? Let $X\mid Y\thicksim\operatorname{Binomial}(Y,p)$, and let $Y\thicksim\operatorname{Binomial}(n,q)$. I have seen it written that:

Claim: $X$ is marginally $\operatorname{Binomial}(n,pq)$.

There is a simple justification: first, $Y$ members of the population are chosen at random with probability $q$, and then each of these is selected with probability $p$, so that each member is randomly selected in the end with probability $pq$. That is satisfying.
However, my computation of the marginal variance of $X$ disagrees with the answer that should be true if the claim above is true. If the claim above is true, we should have $\operatorname{var} X = npq(1-pq)$. However, I'm getting $$\begin{aligned}
\operatorname{var} X
&= E[\operatorname{var}(X\mid Y)] + \operatorname{var}(E[X\mid Y])\\
&=E[Yp(1-p)] + \operatorname{var}(pY)\\
&=np(1-p) + p^2 n q (1-q).
\end{aligned}$$
This does not equal $npq(1-pq)$. To see this, plug in $p=0.5,q=0.5$, and in that case $npq(1-pq)=0.1875n$ and $np(1-p) + p^2 n q (1-q)=0.3125n$.
I'm wondering if someone can check my math, or point to any resource showing that $X$ is really marginally Bin(n,pq). Thanks.

You have an algebraic error in your working --- since $\mathbb{E}(Y)=nq$ you should have:
$$\begin{align}
\mathbb{V}(X)
&= \mathbb{E}(\mathbb{V}(X\mid Y)) + \mathbb{V}(\mathbb{E}(X\mid Y)) \\[6pt]
&= \mathbb{E}(Yp(1-p)) + \mathbb{V}(pY) \\[6pt]
&= p(1-p) \mathbb{E}(Y) + p^2 \mathbb{V}(Y) \\[6pt]
&= n q p(1-p) + p^2 n q (1-q) \\[6pt]
&= n p q [(1-p) + p (1-q)] \\[6pt]
&= n p q (1 - p q), \\[6pt]
\end{align}$$
which matches the marginal variance of the distribution $\text{Bin}(n,pq)$.
