Probability of getting two events from a Poisson process, given a total of two events

Question

Let $\{X_t\},\{Y_y\},\{Z_t\}$ be independent Poisson processes, of rates $\lambda_1,~\lambda_2\,\lambda_3,$ respectively. Assume that events arrive from all the above processes. Find the probability the next two events (from now) concern the process $\{X_t\}.$

Attempt. Let $t>0$ be the time we get exactly two events, that is $X_t+Y_t+Z_t=2$, where $X_t,~Y_t,~Z_t$ are rv's from the Poisson distribution of parameters $\lambda_1t,~\lambda_2t,\,\lambda_3t,$ respectively. Given $X_t+Y_t+Z_t=2$, the rv $X_t$ is from the Binomial distribution $Bin\Big(2,\frac{\lambda_1}{\lambda_1+ \lambda_2+\lambda_3}\Big)$ (classic result from probability), so the desired probability of $X_t=2$ is $\Big(\frac{\lambda_1}{\lambda_1+ \lambda_2+\lambda_3}\Big)^2.$

Thanks in advance for the tip!

Answer 1

Good job. The answer is indeed

$$\left( \frac{\lambda_1}{\sum_{i=1}^3\lambda_i}\right)^2$$