Assume that you have 2 Independent Poisson processes with rates λ1 and λ2. The probability that the 1st event that will occur will be from the first process in a time period t is $\frac{λ1}{λ1+λ2}$ and from the second $\frac{λ2}{λ1+λ2}$. Those processes can be merged to a single process with rate $L=λ1+λ2$ therefore from poisson pmf the probability of no events will occur from either of the processes in a time period is $e^{-L}$.
- Doesn't that make that the first event will occur from the first process $\frac{λ1}{λ1+λ2}*(1- P(no\ events\ happened)) $ and similarly from the second $\frac{λ2}{λ1+λ2}*(1- P(no\ events\ happened))$?
- If so probability that the nth event that will occur will be from the first process is $$\frac{λ1}{λ1+λ2}*(1- P(no\ events\ happened)-P(1\ event\ happened)-...-P(n-1\ events\ happened))$$
If all the assumptions are true what is the probability that the first process will reach a number of event occurrences x before the second process does and what is the probability that neither of those processes will reach x?
I'll provide you with two different approaches to find the probability that the first processes occurs $x-$times before the second process does.
Let $T_x(j):j=1,2$ be the time at which the $x^{\text{th}}-$ event takes place for the $j^{\text{th}}$ Poisson process. You're tasked to compute $\mathbb{P}\left(T_x(1)<T_x(2)\right)$.
Since $T_x(1),T_x(2)$ are independent Erlang random variables, the joint distribution of the bivariate random vector $\left(T_x(1),T_x(2)\right)$ factors as the following: $$f_{T_x(1),T_x(2)}(u,v)=f_{T_x(1)}(u)f_{T_x(2)}(v)=\frac{(\lambda _1 \lambda _2)^{x}}{\left[(x-1)!\right]^2}(uv)^{x-1}\exp\{-(\lambda_1 u + \lambda_2 v)\}\cdot 1_{[0,\infty)^2}(u,v)$$ So, $$\mathbb{P}\left(T_x(1)<T_x(2)\right)=\int_0^{\infty} \int_0^v \frac{(\lambda _1 \lambda _2)^{x}}{\left[(x-1)!\right]^2}(uv)^{x-1}\exp\{-(\lambda_1 u + \lambda_2 v)\}\mathrm{d}u\mathrm{d}v$$
Another approach: Take $N$ as the total number of events that occur in the compound Poisson process until $x$ occurrence of the first process takes place. You're tasked to compute $\mathbb{P}\left(N=x,\dots ,2x-1\right)$.
Note $N$ is supported on $\{x,x+1,\dots\}$ and, for $n\geq x$ fixed, $$\mathbb{P}(N=n)={n-1 \choose x-1}\left(\frac{\lambda_1}{\lambda_1 + \lambda _2}\right)^{x} \left(\frac{\lambda_2}{\lambda_1 + \lambda _2}\right)^{n-x}$$ Therefore $$\mathbb{P}\left(N=x,...,2x-1\right)=\sum_{n=x}^{2x-1}{n-1 \choose x-1}\left(\frac{\lambda_1}{\lambda_1 + \lambda _2}\right)^{x} \left(\frac{\lambda_2}{\lambda_1 + \lambda _2}\right)^{n-x}$$ For the second question, take $E$ as the event that neither process reaches $x$ total occurrences and $T_x(1)$ as the arrival time of the $x^{\text{th}}$ arrival in the first process. Fix $t>0$ arbitrarily. Then $$\begin{eqnarray*}\mathbb{P}(E) &\leq & \mathbb{P}\left(T_x(1)>t\right) \\ &=& 1- \mathbb{P}(T_x(1)\leq t) \\ &=& 1-F_{T_x(1)}(t)\end{eqnarray*}$$ The above is true for any $t>0$ which implies $\mathbb{P}(E)=0$