Consider a single-server queue where arrivals are Poisson with rate $10$ per hour. An arriving customer, upon finding $n$ in the system, departs the system with probability $p_n = \frac{n}{n+1}$ (so, as the system becomes more congested, arriving customers are more likely to go elsewhere without ever entering the line.) Suppose that the time to service each customer is exponential with rate $10$ per hour.
(a) Find the birth and death rates
(b) Find the expected time for the system to reach $3$ customers starting from the empty system.
My attempt: (a) The birth rate is $10q_n = \frac{10}{n+1}$, and the death rate is $\frac{10n}{n+1}$.
(b) My thought is: I tried forming the recusion formula by letting $E_i =$ expected time to reach $i$ customers. Now, $E_{i} = \frac{10}{10*2+10}E_{i-1} + \frac{20}{20+10}E_{i+1}$ with $E(0) = 0$ and $E(1) = 6$ minutes. Thus $E(2) = 9$ and $E(3) = (9 - 2)\frac{3}{2} = \fbox{$10.5$}$ minutes. Is this correct?
My question: I am not sure at all about my solution above. Could anyone give me some thoughts in case my way to solve this one was completely wrong?
The death rate here, i.e. the rate to go from $n$ to $n-1$ when $n \geq 1$, is just $10$ uniformly. It has nothing to do with the customers that leave the system before properly entering it; they never change the state.
The birth rate here is analogous to what happens in the Metropolis dynamics: you "try" to go from $n$ to $n+1$ with rate $10$ but you only succeed with probability $\frac{1}{n+1}$, which is effectively the same as actually going from $n$ to $n+1$ with rate $\frac{10}{n+1}$ (which is what you said). This is a nontrivial but ultimately elementary fact: the sum of $Geo(p)$-many independent exponential random variables with common rate $\lambda$, where the $Geo(p)$ variable is independent of the exponential variables, is an exponential random variable with rate $p \lambda$.