What is the limiting probability that there are n people in the facility? Customers arrive at a store at a Poisson rate of λ and there is a single server with rate μ. The arrival and service times are independent random variables. Customers leave the facility immediately after receiving the service. Find the limiting probability that there are n people in the facility,n=0, 1, 2,... 
I saw this question in a stochastic book and found a solution but I am not sure.
I thought like P0= λ/ μ
P1= λ/ μ * P0
…
Pn= (λ/ μ)n * P0
Do you think that is right?
 A: This is a $M/M/1$ queue with arrival rate $\lambda$ and service rate $\mu$. Let $X(t)$ be the number of customers in the system at time $t$, then $\{X(t):t\geqslant0\}$ is a continuous-time Markov chain on $\{0,1,\ldots\}$ with generator Q given by
$$
q_{ij} = \begin{cases}
\lambda,& j=i+1\\
\mu,& j=i-1\\
-(\lambda+\mu),& j=i, i>1\\
-\lambda,& j=i=1\\
0,& \text{otherwise}.
\end{cases}
$$
Since this is a birth-death chain ($q_{ij}=0$ when $|i-j|>1$), we have the detailed balance equations
$$\lambda \pi_i = \mu\pi_{i+1},\quad i=0,1,2,\ldots $$
This yields the recurrence $\pi_i = \rho^i \pi_0$ where $\rho=\lambda/\mu$. Assuming that $\rho<1$, from the condition $\sum_{i=0}^\infty \pi_i=1$ that a stationary distribution $\pi$ must satisfy, we have
$$
\pi_0 = \left(\sum_{i=0}^\infty\rho^i \right)^{-1} = 1-\rho.
$$
It follows that the limiting probability $\lim_{t\to\infty}\mathbb P(X(t)=n)$ is given by
$$
\pi_n = \rho^n(1-\rho).
$$
A: You’re close to the right answer!
If $\lambda > \mu$, then the system does not possess a stationary distribution as the expected number of people in the system grows and grows indefinitely. 
If $\lambda < \mu$, you’ll find the long-term probability the queue has $n$ customers, $\pi_n$, is:
$$\pi_n = (1 - \frac{\lambda}{\mu})(\frac{\lambda}{\mu})^n$$
The difference between my answer and yours is only a normalizing factor; both of our answers are invariant measures for the stochastic process, but only the answer above is also a probability distribution. 
