Expected Waiting Time: memoryless property 
Smith is waiting for his two friends Lee and Yang to visit his house. The time until Lee arrives is Exp($\lambda_1$) and the time until Yang arrives is Exp($\lambda_2$). After arrival, Lee stays an amount of time that is Exp($\mu_1$), whereas Yang stays an amount of time that is Exp($\mu_2$).
All four random variables are independent.
What is the expected time of the first departure?
Hint: Let X be the time of the first departure, and write X=F+Y where F is the time of the first arrival and Y is the additional amount of time until the first departure. Compute E(Y) by conditioning on who arrived first.

Attempt:
Denote S=Lee's arrive time; U=Lee's stay time; T=Yang's arrive time; V=Yang's stay time.
$E\{first arrival\}=E[X]=E[\min(S,T)]=\frac 1{\lambda_1+\lambda_2}$
After that, everything refreshes, then the additional waiting time is again:
$E[A]=E[\min(U,V)]=\frac 1{\mu_1+\mu_2}$.
Therefore, total time of expected first departure = $\frac 1{\mu_1+\mu_2}+\frac 1{\lambda_1+\lambda_2}$.
Which is wrong
 A: We have
$$
\mathbb E[A] = \frac1{\lambda_1+\lambda_2}
$$
and 
\begin{align}
\mathbb E[F] &= \mathbb E[F\mid \text{Lee arrives first }]\mathbb P(\text{Lee arrives first})\\ &+ \mathbb E[F\mid \text{Yang arrives first }]\mathbb P(\text{Yang arrives first})\\
&= \left(\frac1{\mu_1+\lambda_2} + \left(\frac{\lambda_2}{\mu_1+\lambda_2}\right)\left(\frac1{\mu_1+\mu_2} \right) \right) \left( \frac{\lambda_1}{\lambda_1+\lambda_2}\right)\\
&+\left(\frac1{\mu_2+\lambda_1} + \left(\frac{\lambda_1}{\mu_2+\lambda_1}\right)\left(\frac1{\mu_1+\mu_2} \right) \right) \left( \frac{\lambda_2}{\lambda_1+\lambda_2}\right).
\end{align}
It follows that
\begin{align}
\mathbb E[X] &= \mathbb E[A] + \mathbb E[F]\\
&= \left(\frac1{\lambda_1+\lambda_2}\right)\left(1 +  \frac{\lambda _1 \left(\lambda _2+\mu _1+\mu _2\right)}{\left(\mu _1+\mu _2\right) \left(\lambda _2+\mu _1\right)} + \frac{\lambda _2 \left(\lambda _1+\mu _1+\mu _2\right)}{\left(\mu _1+\mu _2\right) \left(\lambda _1+\mu _2\right)} \right).
\end{align}
A: Everything doesn't refresh after first arrival.   The first person could depart before the other person even arrives.
Suppose Lee arrives first: On that condition: the additional time until first departure is then the minimum time until: Lee's departure, or Yang's arrival and departure. 
$\begin{align}\mathsf E(X) = & ~\mathsf E[(S+U)\wedge (T+V)]
\\ = & ~ \mathsf E(S\wedge T) + \mathsf P(S<T)~\mathsf E [U\wedge (T+V)]+\mathsf P(S>T)~\mathsf E[(S+U)\wedge V]
\\ = & ~\mathsf E(S\wedge T) + \mathsf P(S<T)~(\mathsf E(U\wedge T)+\mathsf P(U>T)~\mathsf E(U\wedge V))+\mathsf P(S>T)~(\mathsf E(S\wedge V)+\mathsf P(S<V)~\mathsf E(U\wedge V))
\\ = & ~\frac{1}{\lambda_1+\lambda_2} + \frac{\lambda_1}{\lambda_1+\lambda_2}~(\frac{1}{\mu_1+\lambda_2}+\frac{\lambda_2}{\mu_1+\lambda_2}~\frac{1}{\mu_1+\mu_2})+\frac{\lambda_2}{\lambda_1+\lambda_2}~(\frac{1}{\lambda_1+\mu_2}+\frac{\lambda_1}{\lambda_1+\mu_2}~\frac{1}{\mu_1+\mu_2})
\end{align}$
A: Supplement for @Math 1000's results. $\mathbb{E}[\text{F}|\text{Lee arrives first}] = \left(\left(\frac{\mu_1}{\mu_1+\lambda_2}\right)\left(\frac1{\mu_1}\right) + \left(\frac{\lambda_2}{\mu_1+\lambda_2}\right)\left(\frac1{\mu_1+\mu_2} \right) \right) \left( \frac{\lambda_1}{\lambda_1+\lambda_2}\right)$, where $\frac{\mu_1}{\mu_1+\lambda_2}$ is the probability of Lee has left when Yang arrives and $\frac1{\mu_1}$ is the average first leaving time under such the condition. The other part is similar.
