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Rock tickets are sold at a ticket counter. Females and males arrive at times of independent Poisson processes with rates 30 and 20.

What is the probability that the first three customers are female

My Work

Let $F(t), M(t)$ be the number of females, males respectively, that arrive up to time $t$. Then if I condition on $M(t) + F(t) = 3$, I get $$\Bbb P(F(t) = 3, M(t) = 0 | M(t) + F(t) = 3) = {\exp(-30t) (30t)^3/3! \cdot \exp(-20t)(20t)^0/0! \over \exp(-50t)(50t)^3 /3!}$$

Is this accurate? My answer, after canceling, works out to ${30^3 \over 50^3}$ which seems a bit too simple? It's exactly $\Bbb P (\text{female arrival})^3$.

Moderat
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  • Why would that be too simple? This is a memoryless process. – Douglas Zare Oct 31 '12 at 15:01
  • Work through the general case where you want to know $\mathbb{P}(F(t)=k|M(t)+F(t)=n)$. You should find that in the general case the number of females, given the total of males and females, will follow a binomial distribution with probability of success equal to the rate of females over the sum of female and male rates. The same goes for the number of males, but your question is dealing with the females. – assumednormal Oct 31 '12 at 15:14
  • Aha! I see this now. I got lucky because the binomial coefficient in this case is 1. Thank you both – Moderat Oct 31 '12 at 15:24
  • Since each arrival gender is independent (due to the memorylessness Douglas Zare noted), $\Pr(\text{next arrival female})^3$ is exactly what I would expect. – Henry Oct 31 '12 at 15:30

1 Answers1

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Suppose $F$ and $M$ are two Poisson processes with rates $r_f=30$ and $r_m=20$, respectively. $F(t)$ and $M(t)$ give the number of arrivals prior to time $t$ for each of the processes.

Given that the sum of $F(t)$ and $M(t)$ is fixed at $n$, what is the probability of having $k$ females prior to time $t$? In other words, what is $\mathbb{P}(F(t)=k|F(t)+M(t)=n)$?

\begin{eqnarray*} \mathbb{P}(F(t)=k|F(t)+M(t)=n) &=& \frac{\exp(-30t)(30t)^k/k!\times\exp(-20t)(20t)^{n-k}/(n-k)!}{\exp(-50t)(50t)^n/n!}\\ &=& \binom{n}{k}\frac{(30t)^k(20t)^{n-k}}{(50t)^{n}}\\ &=& \binom{n}{k}\left(\frac{30}{50}\right)^k\left(\frac{20}{50}\right)^{n-k}\\ &=& \binom{n}{k}\left(\frac{30}{50}\right)^k\left(1-\frac{30}{50}\right)^{n-k}\\ \end{eqnarray*}

So, $F(t)|(M(t)+F(t)=n)\sim\textrm{Binomial}(n=n,p=r_f/(r_f+r_m))$.

assumednormal
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