Continuous time markov chain for automobile shop Given customers arrive at an automobile shop in a Poisson process with rate $\lambda_1 = 2$ per hour. If there are $\geq 3$ customers, the owner would turn away the arriving customers. These customers then leave. Finally, the times to complete service for customers are independent exponential random variables with mean $1$ per hour.
(a) Find the average number of customers in the shop.
(b) When the shop is full ($3$ customers), find the probability of at least one automobile completes the service before the next one arrives.
(c) Find the fraction of customers turned away.
My attempt I was able to construct the rate transition matrix $Q$ as $$Q =\pmatrix{-2 & 2 &  0 &  0 \\ 1 & -3 &  2 &  0 & \\ 0 &  1 &  -3 &  2 & \\ 0 &  0 &  1 &  -1}$$
(the 1st column/1st row = 0 cars in the shop, 2nd column/2nd row = 1 car in the shop, etc. and the 4th column = $\geq 3$ cars in the shop)
But I could not see how to use this transition matrix Q to solve (b), so could anyone please help with this part?
For part (c), I think this is just $1 - P(\text{0 car is finished before the next car arrives}) = 1 - \frac{1}{2+1} = \frac{2}{3}$. Is this correct though?
For part (d), I need to find $P_3$ for the equation $PQ = 0$ with $\sum_{i=0}^{3}  P_i = 1$. My answer is $\frac{8}{15}$.
My question: Could anyone please help me with part (a)? Also, please let me know if I got the other two parts incorrectly. Really appreciate any inputs.
 A: Let $Z(t)$ be the number of customers in the system at time $t$, then $\{Z(t):t\in\mathbb R_+\}$ is a continuous-time Markov chain on state space $S:=\{0,1,2,3\}$ with generator matrix $Q$ as written in the question. Since $\{Z(t)\}$ is irreducible on a finite state space, it has a unique stationary distribution $\pi$. $\{Z(t)\}$ is a birth-death process, so $\pi$ satisfies the detailed balance equations
\begin{align}
\pi_0\lambda &= \pi_1\mu\\
\pi_1\lambda &= \pi_2\mu\\
\pi_2\lambda &= \pi_3\mu\\
\end{align}
Let $\rho=\frac\lambda\mu(=2)$, then we have $\pi_1=\rho\pi_0$, $\pi_2=\rho^2\pi_0$, and $\pi_3=\rho^3\pi_0$. It follows from $\sum_{i\in S}\pi_i=1$ that 
$$\pi_0(1+2+4+8) = 1$$ and hence $$\pi = \left(\frac1{15},\frac2{15},\frac4{15},\frac8{15}\right). $$
For $(a)$, the limiting mean number of customers in the shop is given by
$$L = \sum_{i\in S}i\pi_i = 1\cdot\frac2{15}+2\cdot\frac4{15}+3\cdot\frac8{15}= \frac{34}{15}.$$
For $(b)$, let $S$ be the number of services completed before the next arrival, then 
$$\mathbb P(S\geqslant 1)=1-\mathbb P(S=0) = 1 - \frac\lambda{\lambda+\mu}= 1 - \frac2{2+1} = \frac13. $$
For $(c)$, the fraction of customers turned away is the fraction of customers that arrive when the shop is full: $\pi_3 = \frac8{15}$.
I didn't see a part $(d)$, but indeed $\pi Q=0$ and $\pi_3=\frac8{15}$, as that part seems to be asking.
