In a shop two employees are working. When we get inside the shop, we see that the two employees are already serving two customers (one customer for each employee), with the service time being a random variable of the exponential distribution of parameter λ (independent for each other). each of employees which finish his works will do our work.
What is the probability that we finish our work and go out of shop after that two customers which were already in shop?
I tried to consider the problem in this way which the wanted probability is equal to probability of difference of service time of two first customers is greater than our service time. probability of service time of one customer is greater than the other one can be obtained as $1/2$ but I don't know how to calculate wanted probability which contains difference of service time of two customers.
any help is appreciated , thanks in advance!
Hint: Use the memoryless property of the exponential: That is, after some time, $a$, has elapsed, the distribution of additional time is the same as the original exponential distribution. And so $E[X|X>a]=a+E[X]$.