In a car repair shop with only one birth to repair the cars, cars arrive for repair in a poison process at a mean rate of four cars per day. The manager decides not to accept any car for repair if he already has three cars in the workshop including the one he is repairing. He is able to repair on average eight cars per day. Assume the repair time of the car has exponential distribution. What is the fraction of customers lost?
Textbook answer is: 1/15
Can someone please explain. I've worked out the probabilities $P_0, P_1, P_2, P_3$ using the model M/M/1:FCFS/3/$\infty $ and got $P_0 = \frac{8}{15}, P_1 = \frac{4}{15}, P_2 = \frac{2}{15}, P_3 = \frac{1}{15}$
I'm intended on thinking that fraction of customers lost is $P_3$ but I don't see the connection.
The PASTA property of a Poisson process says that an arriving customer will see the system in any state $i$ the same proportion of time as the long-term proportion of time the system spends in that state, which for this problem is $P_i$.
A customer is lost if he arrives when the system is in state $3$ and that occurs $P_3=1/15$ of the time so $1/15$ of all customers are lost.