= Solution
The <exponential series> proves <normalization of a probability distribution>:
$$
\sum_{n=0}^\infty e^{-\lambda}\frac{\lambda^n}{n!}=e^{-\lambda}e^\lambda=1.
$$
Let $M$ and $F$ count the two labels. Conditional on $N=M+F=k+l$, independent labeling gives a <binomial distribution>, so the <joint probability mass function> is
$$
\begin{aligned}
\mathbb P(M=k,F=l)&=\mathbb P(N=k+l)\binom{k+l}{k}p^k(1-p)^l\\
&=e^{-\lambda}\frac{(\lambda p)^k}{k!}\frac{[\lambda(1-p)]^l}{l!}\\
&=\left[e^{-\lambda p}\frac{(\lambda p)^k}{k!}\right]\left[e^{-\lambda(1-p)}\frac{[\lambda(1-p)]^l}{l!}\right].
\end{aligned}
$$
Each bracket is a normalized <Poisson distribution>. Summing over $l\geq0$ therefore gives
$$
\boxed{\mathbb P(M=k)=e^{-\lambda p}\frac{(\lambda p)^k}{k!},\qquad M\sim\operatorname{Poisson}(\lambda p).}
$$
The factorization also proves that $M$ and $F$ are <independent random variables>. This is <Poisson thinning>; the phrase about ignoring the other count refers to taking a <marginal distribution>, rather than fixing that count.
Back to article page
