Expected values using a poisson process 
A doctor has appointments at 9 and 9:30. The amount of time each appointment lasts is exponential with mean 30 minutes. What is the expected amount of time after 9:30 until the 2nd patient has completed his appointment?

My idea here is to condition on the first appointment taking more than (and less than) 30 minutes. Thus, where $T$ is the amount of time after 9:30 until both appointments are done, and $t_1$ is the time appointment 1 takes, $$E[T] = E[E[T | t_1]] = E[T | t_1 < 30] * P(t_1 < 30) + E[T | t_1 > 30] * P(t_1 > 30)$$ Breaking it down a bit, we see that $$P(t_1 < 30) = (1 - \frac{1}{e}), P(t_1 > 30) = \frac{1}{e}$$ Further, $$E[T | t_1 < 30] = \frac{1}{\lambda} = 30$$ However, this $E[T | t_1 > 30]$ is giving me problems. I would think we should apply the memoryless property here; that is, given the first appointment has already taken 30 minutes, we expect that the 2 appointments together will take 60 more minutes. That is, $$E[T | t_1 > 30] = 2 * \frac{1}{\lambda} = 60$$ This gives me that the expected amount of minutes past 9:30 until the 2nd appointment is done is approximately 41.04 minutes.
I'm fairly confident in everything other than my calculation of $E[T | t_1 > 30]$. Have I gone wrong here, or is this answer correct?
 A: Why all the craze with conditionings? These mainly seem like occasions of going wrong... so, let us keep things simple and let us start from the fact that you are after the mean of $$T=(t_1-30)^++t_2$$ where $t_i$ is the length of appointment $i$, $t^+=\max(t,0)$, and $t_1$ and $t_2$ are both exponential with parameter $\frac1{30}$. Each $t_i$ is distributed like $30\,t_0$ where $t_0$ is standard exponential, hence $$E(T)=30\,E((t_0-1)^+)+30\,E(t_0)$$ One should definitely know that $E(t_0)=1$, and one can compute 
$$E((t_0-1)^+)=\int_1^\infty (t-1)\,e^{-t}dt=\left.-te^{-t}\right|_1^\infty=e^{-1}$$ Finally, $$E(T)=30\,(1+e^{-1})$$
Nota: Of course, one could also use the fact that $t_0-1$ conditionally on $t_0>1$ is distributed as $t_0$, hence $$E((t_0-1)^+)=E(t_0)\cdot P(t_0>1)=1\cdot e^{-1}$$ but this does not sem like a major simplification.
A: Let's say the appointments take $t_1$ and $t_2$, and $S$ is total time elapsed to the end of the second appointment. Then $T = S - 30$ and
$$
\mathbb{E}[S | t_1 < 30] = 30 + \mathbb{E}[t_2] = 60.
$$
and
$$
\mathbb{E}[S | t_1 > 30]
 = \mathbb{E}[t_1 + t_2 | t_1 > 30]
 = \mathbb{E}[t_1 | t_1 > 30] + 30 > 60.
$$
