Shop customers Poisson process People arrive at some shop as Poisson process of rate $N$. There are $N$ corridors in the shop and each customer chooses one at random, independently of the other customers. Now let $X_t^N$ be the proportion of corridors which remain empty at time $t$ and $T^N$ the time until half of the corridors have at least one customer, then it holds that $X_t^N\rightarrow e^{-t}, T^N\rightarrow \log 2$ for $N\rightarrow \infty$ in probability.
Well I started step by step. Let $X$ be the RV that describes the number customers that arrived, then we know $P(X=k)=\frac{N^k e^{-N}}{N!}$. I denote each customer with a number so $P$(customer i chooses some corridor)$=\frac{1}{N}$. How can I continue?
 A: What follows can be tightened up, but essentially the position is that with $N$ corridors and a rate of $N$, each corridor is filled independently with a rate of $1$, which does not change with $N$.  
So at time $t$, the probability an individual corridor is still empty is $e^{-t}$ and so the probability that that the proportion $\frac{k}{N}$ of the corridors are empty is ${N \choose k}e^{-kt}(1-e^{-t})^{N-k}$.  The expected number of corridors empty is $Ne^{-t}$ and the expected proportion of corridors empty is $e^{-t}$.  The law of large numbers tells you that as $N$ increases without limit, the proportion of corridors empty converges on its expectation, i.e. to $e^{-t}$.
Solving $e^{-t} = \frac12$ gives $t= \log_e 2$, so both the median time at which a particular corridor is filled and the limit of the time of when half the corridors are filled as the number of corridors increases are $\log_e 2$. You could have achieved this result without the previous result, since the sample median converges on the population median as the sample size increases. 
