What is the fraction of customers lost in a finite queue with one server, M/M/1/k? $k =$ four places and $s = 1$ server
$k=4, \lambda=\dfrac 1 {30}$, $\mu=\dfrac 1 {25}$
The steady-state probs are p0 = 0.2786, p1 = 0.2322, p2 = 0.1935, p3 = 0.1612, p4 = 0.1343. The theory says that a client doesn't do the queue if the system is in state 4. Thus the portion of clients won´t join the queue is p4. is that correct?
yes, you are right, Luis. Many people get confused on that one.
The steady state probabilities are the probabilities that a client arrives in the system to find the system at that state.
In particular, $p4$ is the probability that an arrival finds the system at state $4$ and is rejected.
Hence, the answer is simply $p4$.
This is due to PASTA (Poisson Arrivals See Time Averages)
https://en.wikipedia.org/wiki/Arrival_theorem