Three passengers with different destinations board a shared taxi. The times between successive stops are independent exponential random variables of rate $\lambda$. At each stop one passenger leaves the taxi. This takes no time, and no additional passenger enter. A departing passenger walks home, and indepdently of everything else, the time this takes is an exponential random variable of rate $\mu$.
a) Suppose that the last passenger leaves the taxi at time $t$. If one of the first two passengers left the taxi at time $s$ (which of course is less than $t$), what is the probability that passenger is already home by time $t$?
b) Suppose that the last passenger leaves the taxi at time $t$. What is the probability that both of the other passengers are already home by time $t$?
Attempt:
a) This is a poisson process $N(t)$. A passenger leaves at time $s$, and the walk time is ~$Exp$.
$\Rightarrow P(time\leq t-s)=1-e^{-\mu(t-s)}$ .
(Concern: does "one of the two customer leaves at time $s$" factor into my calculation? If so, how should I reflect that?)
b) Condition on $N(t)=3$, the two passengers leave the taxi with probability $3!/t^3$.
For this part, I am really confuse mainly because I am not clear about the concept of conditional arrival times in poisson process.
Your answer for a) is correct. For b), conditioning on the last passenger leaving at $t$ leaves the other two passengers' departure times uniformly distributed over $[0,t]$ (since no matter when they depart, the times in the exponents in the product of the probabilities will add up to the constant sum $t$). Thus the probability that a given one of them is home is
$$ \frac1t\int_0^t\left(1-\mathrm e^{-\mu(t-s)}\right)\mathrm ds=1-\frac1{\mu t}\left(1-\mathrm e^{-\mu t}\right)\;, $$
and the probability that they are both home is the square of that.