In a three-server queueing model where arrivals follow a Poisson process with rate $\lambda$ and service times are exponentially distributed with rate $\mu$ per server, what is the probability that there is exactly $1$ customer in the system at steady state?

• A. $\frac{\lambda }{3\mu }\times (1-\frac{\lambda }{3\mu })$
• B. $\frac{\lambda }{\mu +\lambda }$
• C. $\frac{{\lambda }^1}{1!{\mu }^1}\times (1-\frac{\lambda }{3\mu })$
• D. $\frac{{\lambda }^1{\mu }^2}{2!}$

Answer: (C)

Verified step by step guidance

1. Identify the type of queueing model: This is an M/M/3 queue, where arrivals follow a Poisson process with rate \( \lambda \), and there are 3 servers each with exponential service rate \( \mu \).
2. Recall the steady-state probabilities for an M/M/c queue: The probability of having \( n \) customers in the system, \( P_n \), depends on whether \( n < c \) or \( n \geq c \), where \( c = 3 \) here.
3. For \( n < c \), the formula for \( P_n \) is given by: \[ P_n = \frac{(\lambda / \mu)^n}{n!} P_0 \] where \( P_0 \) is the probability of zero customers in the system.
4. Calculate \( P_0 \) using the normalization condition that the sum of all probabilities equals 1. For an M/M/3 queue, this involves summing probabilities for \( n = 0 \) to \( n = 2 \) using the formula above, and for \( n \geq 3 \) using the formula for \( P_n \) when \( n \geq c \).
5. Once \( P_0 \) is found, substitute \( n = 1 \) into the formula for \( P_n \) when \( n < c \) to find the probability of exactly one customer in the system: \[ P_1 = \frac{(\lambda / \mu)^1}{1!} P_0 = \frac{\lambda}{\mu} P_0 \]