# Poisson process different type of events

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??

• In your calculation, did you convert the $20$ minutes to hours? – André Nicolas May 24 '14 at 0:49
• @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? – Danny May 24 '14 at 0:55
• 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. – Deepak May 24 '14 at 0:55
• @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. – Deepak May 24 '14 at 1:00
• @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. – Did May 25 '14 at 8:35