Here is an exercise.
Suppose that people arrive at a bus stop in accordance with a Poisson process with rate $\lambda$. The bus departs at time $t$. Let $X$ denote the total amount of waiting time of all those who get on the bus at time $t$. Let $N(t)$ denote the number of arrivals by time $t$.
Now we want to calculate $E[X\;|\;N(t)]$.
The solution says that the waiting time of each person, $T$, is uniformly distribute in $(0,t)$. So $E[X\;|\;N(t)=n]=n\int_0^t(t-s)\frac{1}{t}ds$.
My question is, why $T$ is uniformly distribute? If people arrivals obeying Poisson process, shouldn't $T$ obeying exponential distribute with parameter $\lambda$?