# Waiting time for two independent poisson processes

Order of Events in Poisson Processes

Assume that you have two independent Poisson process, $N_1(t)$ with rate $\lambda_1$ and $N_2(t)$ with rate $\lambda_2$. The probability that $n$ events occur in the first process before $m$ events occur in the second process is $$\sum_{k = n}^{n+m-1}\binom{n+m-1}{k}\left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^k\left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{n+m-1-k}.$$

Source.

I'm trying to figure out what the expected amount of time would be for at least one event to happen for both processes. I know that the waiting time for Poisson Process in general has an exponential distribution, but I'm not sure how to find the waiting time for two distributions.

Let the inter arrival times be represented by $S\sim\mathcal{Exp}(\lambda)$ and $T\sim\mathcal{Exp}(\mu)$
The probability that at least one of each Poisson process occurs during some interval of length $t$ is:
\begin{align}\mathsf P(\max(S,T)\leq t) & = \mathsf P(S\leq t)~\mathsf P(T\leq t) \\[1ex] & = (1-\mathsf e^{-\lambda t})(1-\mathsf e^{-\mu t}) \end{align}