I know that if $X\sim \mathrm{Poisson}(\lambda)$ and $Y\mid X \sim \mathrm{Bin}(X,p)$ then $Y\sim \mathrm{Poisson}(\lambda p)$. Is there an analogous case for a conditioned Bernoulli distribution? i.e. a problem like:
"The number of people walking into a room during a single day is distributed $\mathrm{Poisson}(\lambda)$. Each person has an (independent) probability $p$ to eat a piece of cake on the table (If it not already eaten). What is the probability the cake is eaten at the end of a given day?"
Edit - Answer (Still happy to receive feedback) : We can define $Y$ (For the stated question) as the number of people who would have eaten the cake (Giving us $Y\sim \mathrm{Poisson}(\lambda p)$ and we have $$\mathbb{P}\left(\text{"The cake was eaten"}\right)=\mathbb{P}\left(Y\geq1\right)=1-\mathbb{P}\left(Y=0\right)=1-\frac{\left(\lambda p\right)^{0}e^{-\lambda p}}{0!}=1-e^{-\lambda p}$$