Question

 At a certain filling station, customers arrive in a Poisson process with an average time of 12 per hour. The time interval between service follows exponential distribution and as such the mean time taken to service to a unit is 2 minutes. Evaluate:
i) Probability that there is no customer at the counter.

ii) Probability that there are more than two customers at the counter.
iii) Average number of customers in a queue waiting for service.
iv) Expected waiting time of a customer in the system.
i) Probability that a customer wait for 0.11 minutes in a queue.

10 Jan 2026
Answer :
Word Count : 281
The arrival rate of customers is (\lambda = 12) per hour, which is equivalent to (\lambda = 12/60 = 0.2) per minute. The service rate is (\mu = 1/2 = 0.5) per minute. This is a standard M/M/1 queue problem. Step 1: Compute traffic intensity [ \rho = \frac{\lambda}{\mu} = \frac{0.2}{0.5} = 0.4 ] i) Probability that there is no customer at the counter For M/M/1 queue, (P_0 = 1 - \rho) [
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