# Waiting time distribution in M/M/1 queuing system?

You wait in a queue served by one assistant. When you enter the queue, the assistant is serving one customer, and another customer is already waiting. The service of one customer has Exponential distribution, with a mean of 1 minute. What is your expected waiting time and what is the distribution of your waiting time?

My approach to this is as follow:

As one is in service and one is in waiting so before our service we will see 2 other services.

Expected waiting time $$= 2*Exp(\mu) = 2-minutes$$

Now how to calculate the distribution of waiting time as we have two exponentially distributed services for two customers and the sum of exponential random variables is not exponentially distributed?

Also I don't have arrival rate $$\lambda$$ here

• You are correct that the sum of two exponentials is not exponential. I suspect the point of this problem is to remind you or get you to figure out what distribution the sum of exponential has... Also note that the sum of two exponential is not the same as 2 times one exponential. – LarrySnyder610 May 12 at 14:05
• @LarrySnyder610 yes, I am guessing the same and I guess I there is a formula for calculation but I am unable to find. – Tigereno May 12 at 14:12
• Try looking online for “sum of exponential random variables” or something similar. – LarrySnyder610 May 12 at 14:12