time for n or more than n customers in a M/M/1 queue system

Question

The question asks for how much time (given an 8hr working day) in a day are there 2 or more customers in the system. If I am to calculate probability for 2 customers in the system and probability for more than 2 customers in the system and then add them up and multiply it by 8, will that give me the answer? Values for lambda and mu are given.

Answer 1

I guess you assume that the system is stable during those 8 hours. Let $\pi_n$ be the stationary probability of having $n$ customers in the system. Then, $\pi_n$ is also equal to the fraction of time you spent in a state with $n$ customers. So, using this knowledge, we find that the fraction of time with 2 or more customers in the system is

\begin{align} \mathbb{P}(Q \ge 2) = 1 - \mathbb{P}(Q<2) = 1 - \mathbb{P}(Q = 1) - \mathbb{P}(Q = 0). \end{align}

You can probably take it from here.