I'm trying to calculate the probability of two poisson processes events happening one before the other, with two different $\lambda$s. The way I see it, I can word it as the probability of event $A$ happening within time $t$ and zero instances of event $B$ happening in the same time $t$.
What i'm confused about though is how to go about writing the probabilities. Specifically, I think what i'm supposed to write is
$P(N(t) < 1)* P(N(t) = 0)$
Where
$P(N(t) < s) = \sum_{i=0}^s e^{-t\lambda}\frac{(t\lambda)^i}{t!}$
I'm not sure if this is correct.
Recall that the arrival times in a Poisson process with rate $\lambda$ follow an Exponential distribution with mean $1/\lambda$. Since inter-arrival times are independent, and you have two Poisson process, then it sounds like you are asking for $$P(T<S)$$ where $T = X_1+X_2$ is the sum of two waiting times for one process, and $S = Y_1+Y_2$ is the sum of two waiting times for the other process. Notice that if the respective rates are $\lambda$ and $\mu$, then $$T\sim\text{Gamma}(2, \lambda)\quad\text{and}\quad S\sim\text{Gamma}(2,\mu).$$
So, you want to compute $$P(T<S) = \int_0^\infty\int_0^s f_T(t)f_S(s)\,dt\,ds = \int_0^\infty\int_0^s \frac{1}{\Gamma(2)}\lambda^2te^{-\lambda t}\cdot\frac{1}{\Gamma(2)}\mu^2se^{-\mu s}\,dtds$$ where the joint density is the product of the densities of $T$ and $S$ since they are independent.
Afterwards, I encourage you to reverse engineer an answer using the binomial distribution and try to see the connection.