This was in my exam today and I'm not sure what's the correct answer.
Let's say that the number of people that enter into a store in the interval $(0,t]$ (in hours) is a Poisson process where $30$ people enter per hour. What is the probability of only $1$ person entering the store between the minute $0$ and the minute $6$, given that $0$ persons enter the store between the minute $0$ and $4$?
I thought of solving this using the variables of the inter-arrival times, which in a Poisson process $X(t) \sim \text{Po}(30t)$ are $T_i \sim \text{Exp}(30)$. Since $T_1$ is the time between $0$ and the first event, I formulated this as follows:
$\Pr(\frac6{60}\geq T_1>0\mid T_1 > \frac4{60})$
This sounded correct to me, because we want to find the probability that the first event lands between the $0$th and $6$th minute, given that the first event did not happen before the $4$th minute.
The calculations I did go as such:
$\Pr(\frac6{60}\geq T_1>0\mid T_1 > \frac4{60})=\frac{\Pr(\frac6{60}\geq T_1 > 0,T_1 > \frac4{60})}{\Pr(T_1 > \frac4{60})}$
the intersection in the numerator is $\Pr(\frac6{60}\geq T_1 > \frac4{60})$. Knowing $T_1$ is exponential with parameter $30$, this gives a probability of $e^{-2}-e^{-3}$. The denominator is just $e^{-2}$, so the probability is approximately $0.63212$.
Is this correct? I know I made a stupid error calculating $\Pr(T_1 > \frac4{60})$ where I got $1-e^{-2}$ instead of $e^{-2}$, but I think my reasoning should be sound...
It's simpler than that. The probability that one person enters between $0$ and $6$ given that no people enter between $0$ and $4$ is precisely the probability that one person enters between $4$ and $6$ given that no people enter between $0$ and $4$. But these two events are independent.