Customers arrive at a barbers shop at the incidents of a Poisson process of rate λ. Each person is served in order of arrival (by the single barber), and takes an exponential, rate μ service time. Services and arrivals are independent, and λ < μ. (a) In equilibrium, how likely is the shop to be empty? (b) Now suppose arrivals are put off by long queues, so that if there are n customers in the shop then a potential arrival decides not to enter with probability n/(n+ 1). What is the equilibrium probability that the shop is empty now?
A) I get that the invariant distribution is $\pi_k = (1-(\lambda/\mu))(\lambda/\mu)^k$ therefore probability the shop is empty is $\pi_0 = 1-(\lambda/\mu)$
For b I changed the arrivals rate to $\lambda n /(n+1) $ and then I solve $\pi Q = 0 $ to get $\pi_k = \lambda^k \pi_0/(k\mu^k)$
I'm not sure if this is right, and if it is I cannot solve for $\pi_0$