Let $N_1(t),~N_2(t),~N_3(t)$ be independent Poisson processes of rates $\lambda_1,~\lambda_2,~\lambda_3>0$, respectively. Evaluate the probability between two successive arrivals of $N_1$ we get at least one arrival of $N_2$ and at least one arrival of $N_3$.
Attempt. The probability of getting an arrival of $N_i$ (and not of $N_j,~j\neq i$) is $\frac{\lambda_i}{\lambda_1+\lambda_2+\lambda_3}.$ If we already have an arrival of $N_1$, then for
$E_i$= the event we get at least one arrival of $N_i$,
we seek $P(E_2\cap E_3)=P(E_2)P(E_3).$ So:
$$P(E_2)=1-P(E_2^c)=1-\sum_{k=0}^{\infty}\bigg(\frac{\lambda_3}{\lambda_1+\lambda_2+\lambda_3}\bigg)^k\frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}$$
and similar we work for $E_3.$ Am I on the right path?
Thanks in advance.
Best Answer
After the initial $N_1$ you need to either first get $N_2$ and then $N_3$ before $N_1$, or first $N_3$ and then $N_2$ before $N_1$. Thus the probability is
$$ \frac{\lambda_2}{\lambda_1+\lambda_2+\lambda_3}\frac{\lambda_3}{\lambda_1+\lambda_3}+\frac{\lambda_3}{\lambda_1+\lambda_2+\lambda_3}\frac{\lambda_2}{\lambda_1+\lambda_2}=\frac{\lambda_2\lambda_3}{\lambda_1+\lambda_2+\lambda_3}\left(\frac1{\lambda_1+\lambda_3}+\frac1{\lambda_1+\lambda_2}\right)\;. $$