Queuing Theory m/m/1 system

Question

2. A fast-food restaurant has one drive-through window. An average of 40 customers per hour arrives at the window. It takes an average of 1 minute to serve a customer. Assume that interarrival and service times are exponential.

   Q.What is the probability that a customer will have to wait before being served?

Answer 1

I suggest having a read of the book Stochastic Networks by Frank Kelly and Elena Yudovina. In particular, you can find a section titled "$M/M/1$ Queues" on page 22 of the book. There the invariant distribution is determined: writing $\lambda = 40/60 = 2/3 < 1$ for the arrival rate (with departure rate $1$), it is given by $$ \pi(j) = (1 - \lambda) \cdot \lambda^j \quad\text{for}\quad j = 0,1,2,... \, .$$ This is valid if and only if $\lambda < 1$. (For $\lambda > 1$ one can show that no invariant distribution exists: there is a unique invariant measure, up to scaling, but this measure is not summable.)

Now, once we know the invariant distribution, we're done: "customer wait" equals "queue non-empty". So, in equilibrium, $$ P(\text{customer does not have to wait}) = P(\text{queue empty}) = \pi(0) = 1 - \lambda, $$ and hence your probability in question is $\lambda = 2/3$.