91Ó°ÊÓ

Since the special customer spends an exponential amount of time with mean \(1/\theta\) out of the system, we know that the time between arrivals of this special customer follows an exponential distribution with a rate of \(\theta\). Therefore, the average arrival rate of the special customer is \(\theta\).

Let's define the state space as follows: State \(n\) represents the number of ordinary customers in the system, from \(n = 0, 1, 2, ...\) Now, let's set up balance equations for this system. For \(n = 0\) (state 0), the balance equation becomes: Balance In: \(q_{1}(\lambda + \theta) = q_{0}(\mu + \theta)\) Balance Out: \(q_{0}\lambda = q_{1}(\mu + \mu_1)\) For \(n \geq 1\) (state n), the balance equation becomes: \(q_{n-1}\mu + q_{n+1}(\lambda+\theta) = q_{n}(\mu_n+\lambda+\theta)\) Where \(\mu_n = \mu + \mu_1\) if the special customer is in service with the ordinary customer at state \(n\).

Let \(P_{n}\) be the probability that an ordinary customer is bumped \(n\) times. When a special customer arrives, she bumps the ordinary customer with a probability of \(\frac{\mu_1}{\mu+\mu_1}\). Using the balance equations from step 2, we can write the probability \(P_{n}\) as: \(P_{n} = \frac{\mu_1(\lambda+\theta)}{(\mu+\mu_1)(\mu+\theta)} P_{n-1}\) Now, to find the probability \(P_{n}\), we will use the initial condition \(P_{0} = 1\) and iterate the above equation as follows: \(P_{n} = \left(\frac{\mu_1(\lambda+\theta)}{(\mu+\mu_1)(\mu+\theta)}\right)^{n} P_{0}\) Therefore, the probability of an ordinary customer being bumped \(n\) times is: \(P_{n} = \left(\frac{\mu_1(\lambda+\theta)}{(\mu+\mu_1)(\mu+\theta)}\right)^{n}\)