Suppose we have two bank tellers.
The time between customer arrivals for the first bank teller is distributed with an exponential distribution with $\lambda_1$. The time between customer arrivals for the second bank teller is distributed with an exponential distribution with $\lambda_2$.
The first teller defers a customer he gets to the supervisor with probability $p_1$ and the second teller defers a customer he gets to the supervisor with probability $p_2$.
In this setup, I am wondering what is the distribution and expected value between customers deferred to the supervisor.
The number of customers arriving at the first teller in a unit of time, $n_1$, is Poisson distributed:
$$P(n_1=k|\lambda_1)=\frac{\lambda_1^k}{k!e^{\lambda_1}}$$
If the customers are deferred by the first teller to the supervisor independently with a probability $p_1$, then the number of those customers deferred to the supervisor in the same unit time, $m_1$, is distributed according to the binomial distribution:
$$P(m_1=k|n_1,p_1)=\binom{k}{n_1}p_1^k(1-p_1)^{n_1-k}$$
Condition $m_1$ on $\lambda_1$, $p_1$ by marginalizing over $n_1$:
$$P(m_1=k|\lambda_1,p_1)=\sum_{n_1=k}^\infty\frac{n_1!p_1^k(1-p_1)^{n_1-k}\lambda_1^{n_1}}{k!(n_1-k)!n_1!e^{\lambda_1}}$$ $$=\frac{(p_1\lambda_1)^k}{k!e^{\lambda_1}}\sum_{n_1=k}^\infty\frac{(\lambda_1-p_1\lambda_1)^{n_1-k}}{(n_1-k)!}$$ $$=\frac{(p_1\lambda_1)^k}{k!e^{\lambda_1}}e^{\lambda_1-p_1\lambda_1}$$ $$=\frac{(p_1\lambda_1)^k}{k!e^{p_1\lambda_1}}$$ $m_1$ is Poisson distributed with rate $p_1\lambda_1$. Similarly, $m_2$ is Poisson distributed with rate $p_2\lambda_2$, and $m_1+m_2\sim\text{Pois}(p_1\lambda_1+p_2\lambda_2)$. The time between customers being deferred to the supervisor by the two tellers is exponentially distributed with mean $(p_1\lambda_1+p_2\lambda_2)^{-1}$.