After owing a lot to you guys on this website, I finally took my exam on Stochastic Process today! There were some questions in this exam that I was not sure about. I want to share it with you guys and see what you guys think.
This is from my recollection of the exam that I took few hours ago so might not be 100% rigorous. Please understand and let me know if I am missing some details necessary. Of course, for legal issue, this is not completely the same as the exam itself.
Black birds arrive at the John's office at the according to Poisson Process with parameter $\alpha$. Independently of the black birds, yellow birds arrive at the office according to Poisson Process with parameter $\beta$. Starting from 9:00AM until 10:AM, given that there were 3 black birds and 5 yellow birds arrived at the office, what is the probability that all 3 black birds arrived before 5 yellow birds?
My answer:
Unfortunately, I could not solve this question. But I tried my best to guess and gave two solutions:
[Solution 1]
Let $X_i$ ~ Unif($0,1$), $i = 1,2,3$ be the arrival time of the black birds and $Y_k$ ~ Unif($0,1$), $k = 1,2,3,4,5$ be the arrival time of the yellow birds.
Therefore, we calculate:
$\int^{1}_{0} P(max{X_i}<t) \cdot P(min{Y_k}>t) \ dt$
= $\int^{1}_{0} t^3(1-t)^5 \ dt$
Then I could not solve this integral by substitution or integration by parts.
[Solution 2]
$P$($3$ black birds arrived before all $5$ yellow birds) =
$P$(first arrival was black bird) + $P$(second arrival was black bird) + $P$(third arrival was black bird) =
$\ {3 \choose 1}\frac{\alpha}{3\alpha + 5 \beta} + {2 \choose 1}\frac{\alpha}{2\alpha + 5 \beta} + {1 \choose 1}\frac{\alpha}{\alpha + 5 \beta} $
Is any of the solutions correct? Haha :')