0
$\begingroup$

This question already has an answer here:

A certain police officer stops cars for speeding. The number of red sports cars she stops in one hour is a Poisson process with rate 4, while the number of other cars she stops is a Poisson process with rate 1. Assume that these two processes are independent of each other. Find the probability that this police officer stops at least 2 ordinary cars before she stops 3 red sports cars.

How do you solve it by binomial or by poisson? My initial response was to consider p = 1/5

How do you solve poisson process problems in which you need to calculate the probability of one event before another? In this case, the probability of 2 ordinary cars before 3 sports cars. By poisson, I tried calculate the event of 2 ordinary cars and event of 3 red sports cars. What I can't understand is one event before another logic. Can we do these problems by binomial?

$\endgroup$

marked as duplicate by whuber probability Sep 10 '18 at 14:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Please see our help center in relation to homework-style questions (about halfway down the page). $\endgroup$ – Glen_b Sep 9 '18 at 23:05
  • $\begingroup$ @Glen_b this isn't homework. It is for self-study. I just want to know how to solve problems of poisson process if one event happens before another? $\endgroup$ – user218970 Sep 9 '18 at 23:06
  • $\begingroup$ It doesn't matter whether it's specifically set as homework, the requirements are the same. Information about what you want to know belongs in your question rather than in comments. Please edit your question to fit with the information I pointed to. $\endgroup$ – Glen_b Sep 9 '18 at 23:07
  • $\begingroup$ I want to know the general idea behind these problems. I don't even require you to solve it. Just put me in the right direction. I have edited the question as well. I'm new on this site, thank you for guiding me. Would be really grateful. $\endgroup$ – user218970 Sep 9 '18 at 23:08
  • 1
    $\begingroup$ Do you mean "the probability that this police officer stops at least 2 ordinary cars before she stops 3 red sports cars" (1) during a particular hour or (2) in however many hours it takes to stop at least 2 ordinary cars? $\endgroup$ – Alvaro Fuentes Sep 10 '18 at 7:56
0
$\begingroup$

I was thinking that you can try to list all of the outcomes covered by the event "this police officer stops at least 2 ordinary cars before she stops 3 red sports cars" if this event happens within one hour:

2 ordinary and 0 red in the first hour (2,0)

2 ordinary and 1 red in the first hour (2,1)

2 ordinary and 2 red in the first hour (2,2)

3 ordinary and 0 red in the first hour (3,0)

3 ordinary and 1 red in the first hour (3,1)

etc...

within two hours:

1 ordinary and 0 red in the first hour, 1 ordinary and 0 red in the second hour (1,0),(1,0)

1 ordinary and 1 red in the first hour, 1 ordinary and 0 red in the second hour (1,1),(1,0)

etc...

but also:

(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(2,0)

Evidently, there are an infinite number of ways that this event could happen, provided the cop is immortal :)

So, instead of thinking about the individual outcomes (which we will not be able to count), we could think in terms of the sums of these independent $Poisson(\lambda_i)$, which are conveniently distributed as $Poisson(\sum^{}_{} \lambda_i)$.

Let's call the ordinary-car process $O\sim Poisson(1)$ and the red-car process $R\sim Poisson(4)$. For the case of two hours, we are looking for:

$[1 - F_{\sum^{2}_{i=1} O_i}(2)]\times[F_{\sum^{2}_{i=1} R_i}(3)]=[1 - F_{Poisson(1+1)}(2)]\times[F_{Poisson(4+4)}(3)]$

where $F()$ is a c.d.f.

I think this is a good start, but I'm not sure what will happen as you add together more and more Poisson random variables (for 3 hours, 4 hours...). Maybe this helps?

$\endgroup$