We have 3D printer which produce white balls according to the Poisson process with tempo $M$ balls per hour. We let the printer work for a half an hour then we put all the balls together with a black ball in an urn. Then we let the printer work half an hour again and put the balls in the same urn.
If I choose a ball randomly out of that urn what is the probability that I choose the black ball?
Let $B$ be the event that we draw the black ball, and let $W$ be the number of white balls produced in one hour (note that if we let $W_{1}$ be the number of white balls produced in the first half hour, and similarly for $W_{2},$ then $W=W_{1}+W_{2},$ but since this is a Poisson process, which has independent increments, $W_{1},W_{2}\overset{\text{iid}}{\sim}\text{Poisson}(M/2),$ so $W\sim\text{Poisson}(M)$). Then to compute $P(B),$ we condition on $W:$ $$P(B)=\sum_{k=0}^{\infty}P(B|W=k)P(W=k)=\sum_{k=0}^{\infty}\frac{1}{k+1}\left(\frac{M^{k}e^{-M}}{k!}\right)=\frac{1}{M}\sum_{k=1}^{\infty}\frac{M^{k}e^{-M}}{k!}.$$ This yields $(1-e^{-M})/M,$ since $\sum_{k=0}^{\infty}(M^{k}e^{-M})/k!=1.$