The problem is
Customers arrive at a tool crib according to a Poisson process of rate $\lambda = 5$ per hour. There is a single tool crib employee, and the individual service times are exponentially distributed with a mean service time of 10 minutes. In the long run, what is the probability that two or more workers are at the tool crib being served or waiting to be served?
It is given that $\lambda = 5$, and I believe that $\mu = 6$ customers per hour. Then, I think the answer is $$1-\frac{1}{6}-\frac{1}{6}\left(\frac{5}{6}\right)-\frac{1}{6}\left(\frac{5}{6}\right)^2 = \frac{125}{216}.$$ However, the answer in the back is $\frac{25}{36}$.
I was wondering what I did wrong. Or is the answer in the back wrong? Thanks.
We have a $M/M/1$ queue, which is a birth-death process with rates $\lambda=5,\mu=6$.
We want the limiting distribution of the number of workers in the tool crib:
$$\pi_k := \lim_{t\to\infty}P(\text{$k$ workers in the tool crib at time $t$}).$$
Using the formulas for this distribution, the probability we require is
\begin{align} 1-(\pi_0+\pi_1) &= 1-\dfrac{1+\lambda/\mu}{\sum_{k=0}^{\infty}\left(\lambda/\mu\right)^k} \\ & \\ &= 1-\dfrac{1+\lambda/\mu}{1/(1-\lambda/\mu)} \\ & \\ &= 1-\dfrac{1+\lambda/\mu}{1/(1-\lambda/\mu)} \\ & \\ &= \left(\dfrac{\lambda}{\mu}\right)^2 \\ & \\ &= \dfrac{25}{36}. \end{align}