If we have two Poisson process $A$ and $B$ with arrival rate $\lambda_{1}$ and $\lambda_{2}$. Now I have question for calculating the probability of an arrival e.g., from process A in the case of merged Poisson process. Which of the approach we should use.
Considering binomial distribution, given an arrival, the probability that this arrival belongs to process A is $\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}$
Or should we use the traditional Poisson pmf to compute the probability of an arrival from process A, such as using, $$p_{m}=\frac{(\lambda_{1}t)^{m}.e^{-\lambda_{1}t}}{m!}$$ and we can put m=1 to get the probability of an arrival from process A using only $\lambda_{1}$ in the equation
If you have two independent Poisson processes, then the probabiliy that any particular arrival in the merged process was generated by process $A$ is indeed $\lambda_1/(\lambda_1+\lambda_2)$ .
A Poisson process consists of point events (arrivals) that occur over an interval at a constant average rate but each independently of any other arrival in the process.
Thus in the merged process you have arrivals that occur at a constant joint rate $(\lambda_1+\lambda_2)$ and each independently generated by one from the two independent Possion processes.
Your aproach seems to be asking the condtional probability that, in a very small interval, we get one arrival from process A and no arrival from process B, given that we get only one arrival from the merged process in that very small interval. Well...sure.
$$\lim\limits_{\Delta t\to 0^+}\cfrac{~\cfrac{(\lambda_1\Delta t)^1e^{-\lambda_1\Delta t}}{1!}\cdot\cfrac{(\lambda_2\Delta t)^0e^{-\lambda_2\Delta t}~}{0!}}{\cfrac{((\lambda_1+\lambda_2)\Delta t)^1e^{-(\lambda_1+\lambda_2)\Delta t}}{1!}}~=~\dfrac{\lambda_1}{\lambda_1+\lambda_2}$$
[Of course, to use this approach you have to first establish that the count of arrivals in the joint process follows a Poisson distribution with the combined average rate.]