You are in a waiting room with $2$ people ahead of you. There are $4$ service desks with a service time that obeys an exponential distribution with a mean of $1$ hour. Currently all service desks are occupied.
What is the probability that all $6$ customers who arrived before you are also ready before you?
The time for all the $4$ customers currently in service to finish is $\exp(1)$ if I'm not mistaken. But beyond that I am fairly lost.
Best Answer
Obviously, for you to get some service, you need three customers to be ready and free up a desk for the two people and a desk for you. Once you start getting service, you are in a "competition" with the other 3 getting service.
Since the distribution is exponential and it is memoryless, it doesn't matter how much time everyone was getting service before that moment. From the moment you sat down, the service time of each of the customers follows an exponential dist. with parameter 1.
Due to symmetry, the probability that your service time will be the longest is $\tfrac{1}{4}$.