# I am having trouble figuring out how many lambda's (births) there are in a given birth-death Markov process problem.

These questions are not for assignment. I am just confused as to how to set up the problem. I also do not need help calculating the problems at hand.

I understand that in a birth and death problem, $$P_{0}=\frac{1}{1+\Sigma_{n=1}^{\infty}\frac{\lambda_{0}\cdot\cdot\cdot\lambda_{}n-1}{\mu_{1}\cdot\cdot\cdot\mu_{n}}}$$

and

$$P_{1}=\frac{\lambda_{0}P_{0}}{\mu_{1}}$$, $$P_{2}=\frac{\lambda_{1}P_{1}}{\mu_{2}}$$, ... etc. where lambdas are the birth rate and mu is the death rate.

My main point of confusion is figuring out how many lambda's there are in the question.

Take this question for example

"A small barbershop, with 1 barber has room for at most 2 customers. Potential customers arrive at a poisson rate of three per hour, and the successive service times are independent exponential random variables with mean $$\frac{1}{4}$$hour."

I believe it makes sense to say that we have two lambda's in this case since we only have room for 2 customers in the barber shop. So with $$\lambda_{0}$$, we are in state 0 (having 0 customers in barber shop, so now we can enter). Similarly, if we are in state 1 ($$\lambda_{1}$$), there is only one customer in there so another customer can enter. Now we have our two. Is this the right way to think about it?

For the next example:

"Potential customers arrive at a full-service, one pump gas station at a poisson rate of 20 cars per hour. Customers only enter if there are no more than two cars (including the one being tended to) at the pump. Suppose the amount of time required to service a car is exponentially distributed with mean of 5 minutes"

My instincts tell me that this is essentially the same problem as the first one, with a different setting. However, when I see the solution, there are 3 lambda's ($$\lambda_{0}, \lambda_{1}, \lambda_{2}$$). But why would we include $$\lambda_{2}$$? Doesn't state 2 mean there are two cars at the station, and if we can only have up to 2 cars, then this car wouldn't even enter.

I am not sure where my logic is failing.