= Solution
A <Poisson distribution> with parameter $\mu$ has <probability generating function> $G_N(s)=e^{\mu(s-1)}$. The preceding part gives $G_F(s)=e^{\mu p(s-1)}$, the <probability generating function> of a <Poisson distribution> with parameter $\mu p$. To prove <independence>, retain both counts:
$$
\mathbb E[s^Ft^S\mid N]=(ps+(1-p)t)^N.
$$
Consequently the joint <probability generating function> factors as
$$
\mathbb E[s^Ft^S]=e^{\mu(ps+(1-p)t-1)}=e^{\mu p(s-1)}e^{\mu(1-p)(t-1)}.
$$
Comparing the coefficients of $s^ft^k$ proves that the joint <probabilities> are the products of the marginal <probabilities>. Hence \b[$F$ and $S$ are independent, with distributions $\operatorname{Poisson}(\mu p)$ and $\operatorname{Poisson}(\mu(1-p))$]. This is <Poisson thinning>, including the degenerate counts when $p=0$ or $p=1$.
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