Arrivals at a bank counter are considered to be Poisson distribution, with an average time of 12 min between two successive arrivals. The time required to serve is distributed exponentially with a mean of 5 min. The probability that an arrival does not have to wait for service:

Concept:

This problem involves a queuing system with Poisson arrivals and exponential service times (M/M/1 queue). The probability that an arrival does not have to wait for service is equal to the probability that the system is idle (i.e., no customers are being served).

Given:

• Average time between arrivals (\(\lambda^{-1}\)) = 12 min → Arrival rate, \(\lambda = \frac{1}{12}\) customers per minute
• Average service time (\(\mu^{-1}\)) = 5 min → Service rate, \(\mu = \frac{1}{5}\) customers per minute

Step 1: Calculate the Traffic Intensity (ρ)

The traffic intensity is the ratio of arrival rate to service rate:

\[ \rho = \frac{\lambda}{\mu} = \frac{\frac{1}{12}}{\frac{1}{5}} = \frac{5}{12} ≈ 0.4167 \]

Step 2: Determine the Probability of System Being Idle (P₀)

For an M/M/1 queue, the probability that the system is idle (no customers in the system) is:

\[ P_0 = 1 - \rho = 1 - \frac{5}{12} = \frac{7}{12} ≈ 0.5833 \]

The probability that an arrival does not have to wait for service is equal to the probability that the system is idle, which is P₀ ≈ 0.5833