Duronto Express arrives at the Bombay Central station according to a Poisson process of rate 3 trains/hour. Local Line trains arrive according to a Poisson process of rate 4 trains/hour.

Conditionally on the event that 8 trains arrive from 9 am to 9:40 am, what is the probability that no trains arrived between 9:10 and 9:20?

So far I used the Superposition Lemma and so the arrival of all the trains at the station is a Poisson process of rate 7 trains/hr

Now I want to find $P(Z(20)-Z(10)=0 | Z(40) = 8)$, where $Z(t) = X(t)+Y(t)$, but I am stuck at this point and don't know how to solve it from here


1 Answer 1


Not going to give away the answer (completely, at least!), however you might consider using the fact that $Z(20) - Z(10) \sim Poisson(\lambda = 7/6)$.

Furthermore, the probability of the event where $Z(40) = 8$ given that there are no arrivals in $[10,20]$ can be obtained from the probabilities of events, say, that there are no arrivals in the interval $[0,10]$ and 8 arrivals in $[20,40]$, or that there is 1 arrival in the interval $[0,10]$ and 7 arrivals in $[20,40]$, etc.

I think you can take it from here?

  • $\begingroup$ can you do this by using the fact that waiting times are uniformly distributed? $\endgroup$
    – bluelagoon
    Nov 23, 2020 at 12:46

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