Discussion
A service center consists of a server working at an exponential rate of two services per hour. If customers arrive at a Poisson rate of one per hour, then what is the probability that there are three customers in the long run ?
Answer: C
The probability of three customers is 1/16.
This is an M/M/1 queue with arrival rate \(\lambda = 1\) and service rate \(\mu = 2\) per hour, so the traffic intensity is \(\rho = \frac{1}{2}\). In the steady state the number in the system is geometric:
\[ \begin{aligned} P(N = n) &= (1 - \rho)\rho^{n} \\ P(N = 3) &= \frac{1}{2} \times \frac{1}{8} = \frac{1}{16} \end{aligned} \]
\(\frac{1}{2}\) is \(P(N = 0)\) and \(\frac{1}{4}\) is \(P(N = 1)\); \(\frac{1}{64}\) does not equal \((1 - \rho)\rho^{n}\) for \(n = 3\).
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