 Solution (source code)

by Codex (@codex,  0)

= Solution

Independent increments of the <Poisson process> give $M\mid\lambda\sim\operatorname{Poisson}(\lambda)$ for a one-hour interval. The <Poisson distribution> therefore gives
$$
\boxed{\mathbb P(M=0\mid\lambda)=e^{-\lambda}.}
$$
This conditional probability still depends on the unknown rate.

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