Consider a queue (1 server only) where arrivals occurs based on a Poisson process with $\lambda > 0$ and the dropout $\mu > 0$. The thing is each arrival can correspond to 1 client with probability $p$ or 2 clients with probability $1-p$.
How to determine equilibrium condition?
The global balance equations $$ \pi_i\sum_{j\in S\setminus\{i\}} q_{ij} = \sum_{j\in S\setminus\{i\}} \pi_jq_{ji} $$ for $i\in S:=\{0,1,2,\ldots\}$ are given by \begin{align} \lambda\pi_0 &=\mu\pi_1\\ (\lambda+\mu)\pi_1 &= \lambda p\pi_0+\mu\pi_2\\ (\lambda+\mu)\pi_n &= \lambda(1-p)\pi_{n-2}+\lambda p\pi_{n-1}+\mu\pi_{n+1},\ n\geqslant 2. \end{align} This recursion yields $$ \pi_n = \frac{\lambda^{n-1}(\lambda+(n-1)(1-p)\mu)}{\mu^n}\pi_0, $$ and from $\sum_{i=0}^\infty \pi_i=1$ it follows that \begin{align} \pi_0 &= \frac{(\lambda -\mu )^2}{\mu (\mu -\lambda p)}\\ \pi_n &= \frac{\lambda^{n-1}(\lambda+(n-1)(1-p)\mu)}{\mu^n}\cdot\frac{(\lambda -\mu )^2}{\mu (\mu -\lambda p)},\ n\geqslant 1. \end{align}