Poisson Process, arrival times

Question

I am pretty stuck on this problem, specifically parts b and c.

For a, here is what I did so far:

Let $M_T$ be the arrival process for men customers. Then $M_T \sim \exp(2)$ Let $W_T$ be the arrival process for women customers. Then $W_T \sim \exp(4)$ Let $C_T$ be the combined arrival process for all customers. Then $C_T \sim \exp(6)$

Then

\begin{align} & P(M_3 = 7, W_3 = 5 \mid C_3 = 12) \\[8pt] = {} & \frac{P(M_3 = 7, W_3 = 5, C_3 = 12)}{P(C_3 = 12)} \\[8pt] = {} & \frac{P(M_3 = 7, W_3 = 5)}{P(C_3 = 12)} \\[8pt] = {} & \frac{P(M_3 = 7) P(W_3 = 5)}{P(C_3 = 12)} \\[8pt] = {} & \frac{\frac{e^{-6}6^7}{7!}\cdot\frac{e^{-12}12^5}{5!}}{\frac{e^{-18}18^{12}}{12!}}=0.0477 \end{align}

I was thinking that if we condition on there being 12 arrivals in a given interval, the number of male customers in this interval would have a binomial distribution but I am unsure what the parameters would be.

For parts b and c, I am really just confused on how to set up the probabilities. I would appreciate any and all help. Thank you!

Answer 1

For (a), your final answer looks good. But it looks like you typed it incorrectly, where everywhere it says $M_3 = 3$, it should be $M_3 = 7$.

For (b), I have seen a similar problem to this with dice. You can "guess" well by thinking of $$P(M_{T^*} = 1 | C_{T^*} = 1)$$ for some time $T^*$ corresponding to the first arrival. You should see it become nice, with the $T^*$ cancelling and you receiving a nice probability distribution.

For (c), one approach based on the one from part (b) could be to do something similar with conditionals, except now you have more possibilities, since the orders could be: $$MMWW, \\ MWMW, \\ MWWM, \\ WMWM, \\ WWMM, \text{ and} \\ WMMW.$$

(other problem similar here, if you get stuck!)

Answer 2

For b, the arrival time of the first man is $\text{Exponential}(2)$ and that of the first woman is $\text{Exponential(4)}$. So you need to find the probability that $\text{Exponential}(2)-\text{Exponential}(4)>0$.

For c, the arrival time of the second man is $\text{Gamma}(2, 2)$ and that of the second woman is $\text{Gamma(2, 4)}$ (this is because the interarrival times are exponentially distrbuted, and the sum of exponential random variables is gamma). So you need to find the probability that $\text{Gamma}(2, 2)-\text{Gamma}(2, 4)>0$.