Poisson process with different service rates

I am currently doing a tutorial question but I think the answer may be wrong.

The question is as follows:

In a certain system, a customer must first be served by server 1 and then by server 2. The service times at server i are exponential with rate $\mu_{i}$ (i = 1; 2). An arrival finding server 1 busy waits in line for that server. Upon completion of service at server 1, a customer either enters service with server 2 if that server is free or else remains with server 1 (blocking any other customer from entering the service) until server 2 is free. Customers depart the system after being served by server 2.

Suppose now that you arrive to find two others in the system, one being served by server 1 and the other by server 2. What is the expected time you spend in the system?

In the answers they denoted $W_{1}$ the time you wait for server 1 to be available.

Consequently they have shown that:

E($W_{1})$ = $\frac{1}{\mu_{1}} + \frac{\mu_{1}}{\mu_{2}(\mu_{1}+\mu_{2})}$

However I think that the answer is:

E($W_{1})$ = $\frac{\mu_{2}}{\mu_{1}(\mu_{1}+\mu_{2})} + \frac{\mu_{1}}{\mu_{2}(\mu_{1}+\mu_{2})}$

There are four parts to this, I think your question concerns the first item:

1) time spent waiting for server 1 which in turn is split into two cases depending on which clerk finished first: $\frac{1}{\mu_1}\left( \frac{\mu_2}{\mu_1 + \mu_2} \right) + \left( \frac{1}{\mu_2} + \frac{1}{\mu_1} \right) \left( \frac{\mu_1}{\mu_1 + \mu_2} \right)$

But after some algebra, this is equal to $= \frac{1}{\mu_1} \left( \frac{\mu_1 + \mu_2}{\mu_1 + \mu_2} \right) + \frac{\mu_1}{\mu_2(\mu_1 + \mu_2)} = \frac{1}{\mu_1} + \frac{\mu_1}{\mu_2(\mu_1 + \mu_2)}$

Which means that what "they" have shown you is right after all.

I think the problem with your calculation is that you forgot to add $\frac{1}{\mu_1}$ when server 1 finishes first. Remember that this process is memoryless. So when 1 finishes first, you have waited the time you spent with clerk 1, and on top of that, you will have to wait for the person being served at 2 as it they started anew.

BTW, these should be the other parts of the problem in case anyone cares:

2) Time getting service with server 1: $\frac{1}{\mu_1}$

3) Time spent waiting for server 2: $0\left(\frac{\mu_2}{\mu_1 + \mu_2}\right) + \left( \frac{1}{\mu_2} \right) \left( \frac{\mu_1}{\mu_1 + \mu_2} \right)$

4) Time spent getting service with server 2: $\frac{1}{\mu_2}$

The solution should be the addition of the 4 parts.