Assume you have two Geiger counters. Time gaps between following molecules that arriving to the first counter= (Ti-Ti-1) ~exp(λ1) Time gaps between following molecules that arriving to the second counter= (Tj-Tj-1) ~exp(λ2)
Now assume that over one minute the two counter counted overall sum of n molecules, for both of them together. How is the number of molecules that got to each of them distribute?
We will show, by a slightly ugly but straightforward computation, that the number of particles of the first type has Binomial Distribution, with "$p$" equalnto $\frac{\lambda_1}{\lambda_1+\lambda_2}$. This is actually intuitively obvious. Pick any of the $n$ molecules. It has probability $p=\frac{\lambda_1}{\lambda_1+\lambda_2}$ of being of type $1$, and we have independence. But the detailed calculation is still a useful exercise.
The number $X_1$ of molecules of the first kind arriving in unit time is Poisson, parameter $\lambda_1$. Similarly, the number $X_2$ of molecules of the second kind arriving in unit time is Poisson, parameter $\lambda_2$. Finally, the total number $Y$ of molecules arriving in unit time is Poisson parameter $\lambda_1+\lambda_2$.
We want the probability that $X_1=x_1$ and $X_2=x_2$ given that $Y=n$.
We have $$\Pr(Y=n)=e^{-(\lambda_1+\lambda_2)}\frac{(\lambda_1+\lambda_2)^n}{n!}.\tag{$1$}$$
Let $A_1$ be the event $X_1=x_1$, and $A_2$ the event $X_2=x_2$. Let $B$ be the event $Y=n$. We want $$\Pr(A_1\cap A_2|B).$$ As usual this is $$\frac{\Pr(A_1\cap A_2\cap B ) }{\Pr(B)}.$$
We already computed the above denominator, it is given by Formula $(1)$. We now compute the numerator. It is $$\left(e^{-\lambda_1}\frac{\lambda_1^{x_1}}{x_1!}\right)\left(e^{-\lambda_2}\frac{\lambda_2^{x_2}}{x_2!}\right),\tag{$2$}$$ where $x_1+x_2=n$. Divide. We get a certain amount of cancellation, and find that $$\Pr(A_1\cap A_2|B)=\left(\frac{\lambda_1^{x_1}}{x_1!}\right)\left(\frac{\lambda_2^{x_2}}{x_2!}\right) \left(\frac{n!}{(\lambda_1+\lambda_2)^n} \right).$$ But $n=x_1+x_2$, that is, $x_2=n-x_1$. So the above expression can be rewritten as $$\binom{n}{x_1}\left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^{x_1}\left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{n-x_1}.$$
Let $p=\frac{\lambda_1}{\lambda_1+\lambda_2}$. then the above formula is just $$\binom{n}{x_1}p^{x_1}(1-p)^{n-x_1},$$ which we recognize as the Binomial Distribution.
