Consider a two-server parallel queuing system where customers arrive according to a poisson process with rate λ, and where the service times are exponential with rate μ. moreover, suppose that arrivals finding both servers busy immediately depart without receiving any service (such a customer is said to be lost), whereas those finding at least one free server immediately enter service and then depart when their service is completed. (a) if both servers are presently busy, find the expected time until the next customer enters the system. (b) starting empty, find the expected time until both servers are busy. (c) find the expected time between two successive lost customers.

I would assume that customers arrive at the queue according to the poisson process, and then decide whether to enter the queue or leave as per the rules in the question. for (a) I interpret "enter the system" as "join the queue". The expected time for this will be E (time until there is a free slot) + E (time for someone to arrive once a slot is free). Noting that the additional time taken for someone to arrive once a spot is free is independant of the time that the slot became free (memorylessness property of poisson process) The waiting time of a Poisson (/lambda) is exp (/lambda) with mean / frac{1}{/lambda} E (/text{Time someone enters the system}) = / frac{1}{2/mu} + / frac{1}{/lambda} Your post suggests you already understand where / frac{1}{2/mu} comes from.