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An Internet cafe is reached by three kind of clients: type I, type II, type III, according to (independent) Poisson processes of parameters $\lambda_1,~\lambda_2,~\lambda_3$, respectively. Evaluate the probability that $15$ clients of type I reach the cafe, before $6$ of the other categories reach the same cafe, on the interval $[0,t].$

Attempt. Let $\{N_1(t)\},~\{N_2(t)\},~\{N_3(t)\}$ denote the Poisson processes that express the arrivals of clients of types $I,~II,~III$, respectively.

Then the desired probability is: $$P(N_1(t)=15,~N_2(t)+N_3(t)<6)=e^{-\lambda_1t}\frac{(\lambda_1t)^{15}}{15!}\cdot \sum_{k=0}^{5} e^{-(\lambda_2+\lambda_3)t}\frac{((\lambda_2+\lambda_3)t)^{k}}{k!}.$$

Am I on the right path? Thank you!

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    $\begingroup$ Your calculation is correct (although the wording of the question is confusing and could be interpreted differently). $\endgroup$ – QQQ Aug 25 '16 at 11:34
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Recall that if $X\sim\operatorname{Pois}(\lambda)$ and $Y\sim\operatorname{Pois}(\mu)$ are independent, the distribution of $X$ conditioned on $X+Y=n$ is $\operatorname{Bin}\left(n,\frac\lambda{\lambda+\mu}\right)$. For if $0\leqslant k\leqslant n$, we have \begin{align} \mathbb P(X=k\mid X+Y=n) &= \frac{\mathbb P(X=k,X+Y=n)}{\mathbb P(X+Y=n)}\\ &= \frac{\mathbb P(X=k)\mathbb P(Y=n-k)}{\mathbb P(X+Y=n)}\\ &= \frac{\dfrac{\lambda^k}{k!}e^{-\lambda}\dfrac{\mu^{n-k}}{(n-k)!}e^{-\mu}}{\dfrac{(\lambda+\mu)^n}{n!}e^{-(\lambda+\mu)}}\\ &= \frac{n!}{k!(n-k)!}\frac{\lambda^k\mu^{n-k}}{(\lambda+\mu)^n}\\ &= \binom nk \left(\frac\lambda{\lambda+\mu}\right)^k\left(\frac\mu{\lambda+\mu}\right)^{n-k}. \end{align} In this problem, we have $\lambda=\lambda_1$ and $\mu=\lambda_2+\lambda_3$ (which is valid because the superposition of independent Poisson processes is a Poisson process with intensity the sum of that of the component processes). So the probability that there are exactly $15$ arrivals of type $1$ and less than $6$ arrivals of types $2$ and $3$ is given by \begin{align}\scriptsize \mathbb P\left(N_1(t)=15,N_2(t)+N_3(t)<6\right) &=\scriptsize \mathbb P\left(N_1(t)=15\mid N_2(t)+N_3(t)<6\right)\mathbb P\left(N_2(t)+N_3(t)<6\right)\\&=\scriptsize \left(\sum_{m=0}^5 \mathbb P\left(N_1(t)=15\mid N_2(t)+N_3(t)=m\right)\right)\left(\sum_{m=0}^5 \mathbb P\left(N_2(t)+N_3(t)=m\right) \right)\\ &=\scriptsize \left(\sum_{m=0}^5 \mathbb P\left(N_1(t)=15\mid N_1(t)+N_2(t)+N_3(t)=15+m\right)\right)\left(\sum_{m=0}^5\frac{(\lambda_2 t +\lambda_3 t )^m}{m!}e^{-(\lambda_2 +\lambda_3 )t} \right)\\ &=\scriptsize \left(\sum_{m=0}^5 \binom{15+m}{15}\left(\frac{\lambda_1 t }{\lambda_1 t +\lambda_2 t +\lambda_3 t }\right)^{15}\left(\frac{\lambda_2 t +\lambda_3 t }{\lambda_1 t +\lambda_2 t +\lambda_3 t }\right)^m \right)\left(e^{-(\lambda_2 +\lambda_3 )t}\sum_{m=0}^5\frac{(\lambda_2 t +\lambda_3 t )^m}{m!} \right)\\ &=\scriptsize e^{-(\lambda_2 +\lambda_3 )t}\left(\frac{\lambda_1 t }{\lambda_1 t +\lambda_2 t +\lambda_3 t }\right)^{15}\left(\sum_{m=0}^5 \binom{15+m}{15}\left(\frac{\lambda_2 t +\lambda_3 t }{\lambda_1 t +\lambda_2 t +\lambda_3 t }\right)^m \right)\sum_{m=0}^5\frac{(\lambda_2 t +\lambda_3 t )^m}{m!} . \end{align}

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  • $\begingroup$ Thanks! Since $N_1(t),~N_2(t)+N_3(t)$ are independent, why do you work with conditional probability? The formula $P(N_1(t)=15,~N_2(t)+N_3(t)<6)=P(N_1(t)=15)~P(N_2(t)+N_3(t)<6)$ is not valid? $\endgroup$ – Nikolaos Skout Aug 31 '16 at 6:50
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    $\begingroup$ That is correct (and less complicated). I just wanted to solve the problem a different way :) $\endgroup$ – Math1000 Aug 31 '16 at 7:43

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