Arrivals at a telephone booth are considered to be Poisson with an average time of 10 minutes between one arrival and the next. The length of phone calls is assumed to be distributed exponentially, with a mean of 3 minutes. What is the probability that a person arriving at the booth will have to wait?

  • 3
    $\begingroup$ When's the last time you saw a phone booth? Must be from an old book or very unimaginative author or professor. $\endgroup$ Oct 13, 2019 at 16:57
  • $\begingroup$ I just want to know its answer for better understanding.. $\endgroup$
    – techbyte
    Oct 13, 2019 at 16:58
  • 2
    $\begingroup$ Is this a homework question? If so, 1) you can add a new homework tag and 2) what have you done to solve it and where are you stuck? $\endgroup$
    – EhsanK
    Oct 13, 2019 at 17:32

1 Answer 1


This is a $M/M/1$ queue with Poisson arrival distribution with $\lambda =1/10$ and Exponential service distribution with $\mu=1/3$. The proportion of the time that the system is busy can be calculated by using the following equation: $$\rho=\frac{\lambda}{\mu} = \frac{1/10}{1/3}=0.3$$

which is exactly the proportion of time that a person arriving at the booth should wait while the server (telephone booth) is busy. So for a person arriving to the booth the probability of finding the booth busy is $0.3$.

This document will help you understand the basics of queueing models and Kendall's notation for queues.


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