Let $X$ be the number of customers arriving daily at a local restaurant, acconding to a Poisson distribution of parameter $\lambda>0$. Due to technical reasons, the restaurant can serve only the $5$ of them daily. Find the mean number of customers being served daily at the restaurant.
Attempt. The number $Y$ of customers being served daily at the restaurant is a discrete rv, taking values $0,1,2,3,4,5$. But the probability mass is not purely from the Poisson distribution. How do we determine its mass (to find the expected value)?
Following the hint by @LoveTooNap29, $Y$ is a discrete random variable, whose values are $0,1,\ldots,5$ and the probability mass function of $Y$ is $P(Y=k)=P(X=k),~k=0,1,2,3,4$ and $P(Y=5)=P(X\geq 5)$, so its mean value is $$E(Y)=\sum_{k=0}^{5}k\,P(Y=k).$$