Exponential expected service time question

A customer enters a store where he is serviced by clerk 1, followed by clerk 2, finally by clerk 3. Service time by each of the clerks is an exponentially distributed variable with $\displaystyle \mu_i$, where $\displaystyle i =1,2,3$. Assume you enter the the store and there is a single customer being serviced by clerk 3.

**a)** Find the probability that clerk 3 will still be busy when you move to clerk 2.

**b)** Find the probability that clerk 3 will remain busy when you move to see him.

**c)** Find the total time in the store

Solutions:

**a)** For this I'm thinking that $\displaystyle \frac{1}{\mu_3}>\frac{1}{\mu_1}$ for clerk 3 to be busy when you finish with clerk 1. Therefore the probability that clerk 3 is still busy when you move to clerk 2 would be $\displaystyle P(C_1\leq C_3) = \frac{\mu_1}{\mu_1+\mu_2+\mu_3}$. But this doesn't seem correct.

**b)** You're done with clerk 1 and 2 and you have to wait to see clerk 3. Thus the expected time of clerk 3 would have to be greater then that of clerk 1 and 2 combined so you would have $\displaystyle \frac{1}{\mu_3}>\frac{1}{\mu_1+\mu_2}$ which would imply $\displaystyle P(C_1 \ \cup \ C_2 \leq C_3)= \frac{\mu_1+\mu_2}{\mu_1+\mu_2+\mu_3}$

**c)** For this one I'm assuming it would be the total service time + total waiting time (assuming clerk 3 is still busy when you go see him)

so I would have :

$\displaystyle E[T] = \frac{1}{\mu_1} +\frac{1}{\mu_2} +\frac{1}{\mu_3} +\left( \frac{\mu_1+\mu_2}{\mu_1+\mu_2+\mu_3}\right) \times \left( \frac{1}{\mu_3}\right)$