M/M/3 queue - reducing wait time by adding servers Full question below:
You are the manager of the customer support division in your company. Your
division uses 3 telephone lines operated by 3 separate customer service representatives. A
customer is put on hold if their call arrives while all 3 customer service representatives are
busy serving other customers. You observe that customer calls arrive at a Poisson rate of 5
per hour, and that the length of the customer calls is exponentially distributed. You also
observe that 75% of the time, a customer is not put on hold, while the remaining 25% of
the time, a customer can expected to be put on hold for an average of 12 minutes. You wish
to improve service in the division by making sure that 90% of the time, a customer is not
put on hold, while 10% of the time, a customer can expect to be put on hold for an average
of only 4 minutes. How many telephone lines will you add to your division to achieve your
goal?

So I think the biggest problem here is that I don't know $\mu$.  I do know $\rho=\frac{\lambda}{c\mu}=\frac{5}{3\mu}$ for this problem.  I understand that "time on hold" refers to to time waiting in the queue.  With $W$ is time waiting in the queue, I know:
$$E[W]=\frac{\rho}{\lambda(1-\rho)}P(W>0)$$
With 3 operators, I used the fact that "75% of the time, a customer is not put on hold, while the remaining 25% of
the time, a customer can expected to be put on hold for an average of 12 minutes" to calculate:
$$E[W]=.75(0) + .25(12min)=3min$$
Then using $E[W]$ along with $P(W>0)=.25$, I solved the first equation to find $\mu=\frac{10}{3}$. 
Knowing $u$, I used "90% of the time, a customer is not
put on hold, while 10% of the time, a customer can expect to be put on hold for an average
of only 4 minutes" to find the new $E[W]=1min$ and $P(W>0)=.1$.  
To solve for $c$(number of servers) I again plugged these numbers into the original equation for $E[W]$ and found $c=3.3$. You can obviously only have an integer number of servers, so this would be $c=4$, and minus the original 3 would give the addition of just 1 server as the answer.
Sorry for the long question, but am I doing this right?  I feel like I messed up along the way or made some wrong assumptions (mostly that $E[W] can be calculated from the information in the problem).  
Thanks for looking.
 A: I wished to add this as a comment, but I am not allowed to.
We are dealing with an $M/M/c$ queueing system. You calculated the service rate $\mu$ correctly, but then continue incorrectly.
I'd suggest you look at these lecture notes, page 44, equations (5.1) and (5.3) and increase the number of servers $c$ until we have that the probability to wait $\Pi_W := \mathbb{P}(W > 0) \le 0.1$ and the expected waiting time $\mathbb{E}[W] \le 4$ minutes. 
A: Let $X(t)$ be the number of customers in the system at time $t$ (including those on hold). Then $\{X(t):t\geqslant 0\}$ is a continuous-time Markov chain on $\{0,1,2,\ldots\}$. Assuming $\rho :=\frac\lambda{3\mu}$ we can find the stationary distribution $\pi$ with the global balance equations
\begin{align}
\lambda\pi_0 &= \mu\pi_1\\
\lambda\pi_1 &= 2\mu\pi_2\\
\lambda\pi_n &= 3\mu\pi_n, n\geqslant 2.
\end{align}
This yields $$\begin{align}\pi_1 &= \frac\lambda\mu\pi_0\\ 
\pi_2 &= \left(\frac\lambda\mu\right)\left(\frac\lambda{2\mu}\right)\pi_0\\
\pi_{3+i} &= \rho^i\left(\frac\lambda\mu\right)\left(\frac\lambda{2\mu}\right)\pi_0,\  i=0,1,2,\ldots
\end{align}
$$
From $$\sum_{n=0}^\infty \pi_n = 1$$ we obtain
$$\pi_0 = \left[1 + \left(\frac\lambda\mu\right) + \left(\frac\lambda\mu\right)\left(\frac\lambda{2\mu}\right)+ \frac1{3!}\left(\frac\lambda\mu\right)^3\left(\frac1{1-\rho}\right) \right]^{-1}, $$
and thus the probability of being put on hold is
$$\sum_{i=3}^\infty \pi_i = \frac{\pi_3}{1-\rho} = \frac{\frac1{3!}\left(\frac\lambda\mu\right)^3}{1-\rho}\left[1 + \left(\frac\lambda\mu\right) + \left(\frac\lambda\mu\right)\left(\frac\lambda{2\mu}\right)+ \frac1{3!}\left(\frac\lambda\mu\right)^3\left(\frac1{1-\rho}\right) \right]^{-1}. $$
A standard result in queueing theory (e.g. Halfin, Whitt (1981) is that the probability of delay can be approximated for a large number of servers $n$ by $$HW(\beta) = \frac1{1+\beta\Phi(\beta)/\varphi(\beta)}\tag 1 $$
where $\Phi$ is the CDF of the standard normal distribution, $\varphi$ the PDF of the standard normal distribution, and $\beta>0$ satisfies $$\frac\lambda\mu = n - \beta\sqrt n.\tag 2 $$ To obtain $\mathbb P(\mathrm{Delay})<\frac1{10}$, we find $\beta\approx 1.4202$ from $(1)$. Solving $(2)$ for $n$ and substituting $\beta$, $\lambda$ and $\mu$ yields
$$n\sim\frac5{10/3} + 1.4202\sqrt{10/3} = 4.0929. $$
Therefore we should have at least $\lceil 4.0929\rceil=5$ servers.
Given that the service rate is $\mu=\frac5{144}$ per minute, the expected delay time with $5$ servers is
$$\mathbb E[\mathrm{Delay}]= \mathbb P(\mathrm{Delay})\frac{\rho}{\lambda(1-\rho)} = \frac{72}{65}.  $$
