A job is processed with $\lambda \sim \mu_1$ at Station 1
Station 2 processes the jobs at $\mu_2$ but it has a $0.5$ probability of sending the job back to Station 1 (regardless of the # of times it's been processed).
What is the expected waiting time?
I found the probability of the job being processed $n$ times to be $0.5^n$
Thus the expected service time $\mathbb E[S]$
$$=\sum_1^\infty (1/\mu_1 + 1/\mu_2 )n\times0.5^n$$
Is this correct?
If so, I tried to do it using another method:
Suppose I complete the first 2 stations in $1/\mu_1 + 1/\mu_2$ .Then I have $0.5$ probability to go back to where I started, and the expected time is the same again. Thus:
$$\mathbb E[S]=(1/\mu_1 + 1/\mu_2 + \mathbb E[S])\times 0.5$$
Is this correct?
The time in service is $S$. Denote the time it takes to be processed once by station $1$ and once by station $2$ as $B$. This means that
\begin{equation} \mathbb{E}[B] = \frac{1}{\mu_1} + \frac{1}{\mu_2}. \end{equation}
The probability that a job is processed $n$ times by station $1$ and $2$ is indeed $0.5^n$. So,
\begin{equation} \mathbb{E}[S] = \sum_{n = 1}^\infty n \mathbb{E}[B] 0.5^n = \mathbb{E}[B] \frac{0.5}{(1 - 0.5)^2} = 2 \mathbb{E}[B]. \end{equation}
The other approach is almost correct. $\mathbb{E}[S]$ is at least $\mathbb{E}[B]$ and with probability $0.5$ you need to start over. So,
\begin{equation} \mathbb{E}[S] = \mathbb{E}[B] + 0.5 \mathbb{E}[S] \quad \Rightarrow \mathbb{E}[S] = 2 \mathbb{E}[B]. \end{equation}