Question about the Poisson process A service center consists of two servers, each working at an exponential rate 
of two services per hour. If customers arrive at a Poisson rate of three per hour, then, 
assuming a system capacity of at most three customers,
 What fraction of potential customers enter the system?
I was thinking if I could think of the servers as one big system since the customers don't have a preference one way or another.
 A: Time between arrivals is exponential with mean $1/\lambda=1/3$ hour. Service rate $\mu=2$ per hour. Length of a service time is exponential with mean $1/\mu=0.5$ hours. 
This is an $M/M/2/3$ queue. 2 servers with $N=3$ max capacity which means a waiting room of size $1.$
You can use the total number currently in the system as the state space (as opposed to a vector keeping track of which server is busy.) This gives:
state $= 0$ means no one in system.
state $= 1$ means 1 customer being served, no one waiting
state $= 2$ means 2 customers being served, no one waiting
state $= 3$ means 2 customers being served, 1 waiting.
no other states possible.
We want the stationary distribution. It is a birth and death process so we know the answer from the general case. Or we can use the "balance equations." (Tricky part: departure rate in a state is proportional to the number of busy servers.)
After some algebra, $\pi_0=32/143$ and $\pi_3=27/143.$ Check my work.
Now the probability that an arrival is turned away since the system is full is $\pi_3.$ So then we know the proportion of arrivals that are able to get into the system. (Technical point: this is because the PASTA property holds here.)
A: fraction of potential customers enter the system = 1 - πN
here N is system capacity
= 1 -   π3
= 1-27/143
= 0.811188
similar question with solution   {see question 3}
https://www2.isye.gatech.edu/~ashapiro/publications/MT2-solutions.pdf
