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Suppose that it arrives people to a store according to a poisson process with rate $\lambda = 6$/hour , females arrive with probability $0.6$ and male with $0.4$.

What is the probability that there are 4 male and 3 female persons in the store at time $t = 20 $ minutes

here is my answer,which is wrong, maybe someone can explain why?

I know that I can regard Female arrivals as its own poisson process with rate ($0.6*6$)/hour and Male arrivals with rate ($0.40*6$)/hour. let us denote these two poisson processes as ,female : $\{ N_1(t),t\geq 0 \}$ and male: $\{ N_2(t),t\geq 0 \}$.Furthermore I know these are independent.

So i want this probability : $P\{N_1(20) = 3,N_2(20) =4 \}$ by independence i get $P\{N_1(20) = 3\}*P\{N_2(20) =4 \}$ but this is wrong why??

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    $\begingroup$ In your calculation, did you convert the $20$ minutes to hours? $\endgroup$ – André Nicolas May 24 '14 at 0:49
  • $\begingroup$ @AndréNicolas its very late here i can't look at my calculations right now, but is this the right approach? Am I doing it right? $\endgroup$ – Danny May 24 '14 at 0:55
  • $\begingroup$ I'm not 100% sure, but I don't think the approach is right. Male and female arrivals are not independent events. I would work out the overall probability of 7 arrivals (disregarding gender), and then use the binomial distribution to calculate the probability that 4 of those were males (automatically, 6 become females). Multiply the two to get the required probability. $\endgroup$ – Deepak May 24 '14 at 0:55
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    $\begingroup$ @Danny. Pls see my edit. To my mind, they are not independent in the sense that if you have 7 arrivals and 4 males, you necessarily have 3 females. You can't have any other number of females. That's why my approach looks more correct, but I'm not absolutely sure. $\endgroup$ – Deepak May 24 '14 at 1:00
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    $\begingroup$ @Deepak Which is $e^{-p\lambda}(p\lambda)^4/4!$ times $e^{-(1-p)\lambda}((1-p)\lambda)^3/3!$ with $p=0.4$, hence the independence. $\endgroup$ – Did May 25 '14 at 8:35

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