In a city, there is a central bus station. There are $K$ buses that run simultaneously in the city returning to the bus station from time to time. Once a bus is in the station, it stays there for $c$ time-units. The whole bus-station-to-bus-station cycle lasts for $\lambda$ time-units on average, for any bus (this includes the staying time). I'd like to know the probability that there are between $m$ and $n$ buses in the station at the same time. In particular, I am really interested in the special case when there are at least 2 buses in the station.
Let me see if I'm going in the right direction.
Let $N_i(t)$ be the random variable counting the number of times for the $i$-th bus to return to the station between now and time $t$. If I am not mistaken, we have $N_i(t)\sim Poisson(\lambda)$.
I'm about to calculate this quantity :
$$\mathbb{P}\left(m\leq \sum_{i=1}^K\left[N_i(t+c)-N_i(t)\right]\leq n\right)$$
Or maybe I should introduce $X_i$ the time between two cycles for the $i$-th bus? So then $X_i\sim Exp(\lambda)$. Or should I even go as far as taking the sum of those exponential distributions and then study the Erlang distribution? I'm lost at this point.
Thank you.
Each bus has probability $\frac{c}{\lambda}$ of being in the station at a random time. In the long run if the busses get out of sync, those probabilities can be assumed to be independent. Therefore the probability of exactly $i$ busses being in the station is ${K \choose i} (\frac{c}{\lambda})^i(1 - \frac{c}{\lambda})^{K-i}$.
It quickly follows that at least 2 busses has probability $1 - (1 - \frac{c}{\lambda})^K - K \frac{c}{\lambda} (1 - \frac{c}{\lambda})^{K-1}$.
If the bus schedules have not yet become independent, a wide range of answers are possible.