# Poisson process Continuous -time stochastic processes

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

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.