= Solution
Given $M=m$, there are exactly $m$ independent opportunities to stay awake, each with <probability> $p$. Hence
$$
\boxed{W\mid M=m\sim\operatorname{Binomial}(m,p).}
$$
The standard <binomial distribution> formulas are $\mathbb E[W\mid M]=pM$ and $\operatorname{var}(W\mid M)=Mp(1-p)$. The <Poisson distribution> formulas are $\mathbb E[M]=\lambda$, $\operatorname{var}(M)=\lambda$, and therefore $\mathbb E[M^2]=\lambda^2+\lambda$. By the <law of total expectation>,
$$
\mathbb E[WM]=\mathbb E\bigl[M\mathbb E[W\mid M]\bigr]
=p\mathbb E[M^2]=p(\lambda^2+\lambda),
\qquad \mathbb E[W]=p\lambda.
$$
Using the definition of <covariance>,
$$
\boxed{\operatorname{cov}(W,M)=\mathbb E[WM]-\mathbb E[W]\mathbb E[M]=p\lambda.}
$$
Thus $W$ and $M$ are not independent, even though the separate awake and asleep counts produced by <Poisson thinning> are independent.
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