Consider a single-server queue where the arrivals are Poisson with rate $\lambda = 10$ per hour. The service distribution is exponential with rate $\mu = 5$ per hour. Suppose that customers balk at joining the queue when it is too long. Specifically, when there are n in the system, an arriving customer joins the queue with probability $\frac{1}{1 + n}$. Determine the steady-state probability that there are $n$ in the system
My thought: First, since this single-server queue follows birth-death process, $\ p_n = \frac{\lambda_0\lambda_1\ldots \lambda_{n-1}}{\mu_1\mu_2\ldots \mu_n}p_0$. Now, by Bayes' theorem, $P(\text{(n+1)th customer joins in the queue| there are n people in the system}) = \frac{P(\text{n+1 people in the system})}{P(\text{n people in the system})} = \frac{\lambda_n}{u_{n+1}} = \frac{b_n\lambda}{5} = \frac{1}{n+1}$ (since our system has balking, so $\lambda_{n} = b_{n}\lambda\ = 10b_{n}$ where $b_0 = 1$, $b_n$ is non-increasing). Thus, $b_{n} = \frac{1}{2(n+1)}$ and so $\ \lambda_{n} = \frac{5}{n+1}$.
From here, I could compute the steady-state probability $p_0 = (1+ \sum_{n=1}^{\infty} \frac{\lambda_0\lambda_1\ldots \lambda_{n-1}}{\mu_1\mu_2\ldots \mu_n})^{-1} = (1 + \frac{10}{5} + \frac{10\times 5/2}{5^2} + \frac{10\times 5^2/(2.3)}{5^3} +\ldots )^{-1} = (4+\frac{1}{3} + \frac{1}{3.4} + \frac{1}{3.4.5} +\ldots )^{-1}$.
Therefore, $p_n = \frac{\lambda_0\lambda_1\ldots \lambda_{n-1}}{\mu_1\mu_2\ldots \mu_n}p_0 = \frac{10\times 5/2\times 5/3\ldots \times 5/n}{5^n} = \fbox{$\frac{2p_0}{n!}$}$
My question: Could someone help point out the mistake in my solution above? The reason I think there is a mistake is because I would need $\lambda_0 = 10$, but based on the formula I got above for $\lambda_n$, it would give $5$ instead of $10$ when $n=0$. But I could not see where did I made the mistake in terms of the algebra or formulation. Any help would greatly be appreciated.