Given a Poisson process with parameter $\lambda$, what is the probability $P(T_2 \leq 2 | N(3) = 1)$, where $T_2$ is the interarrival time between the first and second arrivals? In words, given that there is one arrival in the first 3 time units, what is the probability that the second arrival occurs within 2 time units of the first arrival?
My attempt: We want to calculate $P(T_2 \leq 2 | N(3) = 1) = \frac{P(T_2 \leq 2, N(3) = 1)}{P(N(3)=1)}$. I know that $P(N(3)=1) = e^{-3\lambda}*3\lambda$. The first arrival occurs in $[0,3)$, and it must be that $T_1 > 1$ for the events {$T_2 \leq 2, N(3) = 1$} to hold. Since $T_1$ is uniformly distributed on $[0,3)$, $P(T_1 > 1) = \frac{2}{3}$. If $T_1 = t_1$ for $1 \leq t_1 \leq 3$, then $P(T_2 \leq 2, N(3) = 1) = P(N(t_1 - 1) > 0)$. Not quite sure where to proceed from here, so would appreciate any hints!
The joint event $(T_2 \le 2) \cap (N(3) = 1)$ indeed requires that $T_1 > 1$. It is natural to condition this event on the first arrival time, so given $T_1 = t$ for some $1 < t \le 3$, then we need $3-t \le T_2 \le 2$. Hence
$$\begin{align} \Pr[(T_2 \le 2) \cap (N(3) = 1)] &= \int_{t=1}^3 \Pr[3-t \le T_2 \le 2 \mid T_1 = t]f_{T_1}(t) \, dt \\ &= \int_{t=1}^3 (e^{-(3-t)\lambda} - e^{-2\lambda}) \lambda e^{-\lambda t} \, dt \\ &= (2\lambda - 1)e^{-3\lambda} + e^{-5\lambda}. \end{align}$$
So the desired conditional probability is $$\Pr[T_2 \le 2 \mid N(3) = 1] = \frac{(2\lambda - 1)e^{-3\lambda} + e^{-5\lambda}}{3 \lambda e^{-3\lambda}} = \frac{2}{3} - \frac{1 - e^{-2\lambda}}{3\lambda}.$$