# Thread: Conditional probability + Poisson question

1. ## Conditional probability + Poisson question

Hi, I have the following problem... I have solved most of it, but I am stuck on this question :

"B) Given that eight buses arrive, what is the probability that six of them are number 29's?"

Thanks a lot for your help!

The problem and what I have done so far is:

"Number 29 and number 42 buses arrive at a bus stop according to independent Poisson processes each at a rate of four per hour.

A) What is the probability that eight buses arrive between 10 and 11am?
B) Given that eight buses arrive, what is the probability that six of them are number 29's?"

$\displaystyle X$ (buses number $\displaystyle 29$) has Poisson distribution with $\displaystyle μ$ and $\displaystyle Y$ (buses number $\displaystyle 42$) has Poisson with $\displaystyle λ$.

Let $\displaystyle Z=X+Y$, therefore $\displaystyle Z$ will have Poisson distribution with parameter $\displaystyle μ+λ$.

$\displaystyle P(Z=8)=(8^8*e^{-8})/8!=0.1396$"

B:

The solution is supposed to be $\displaystyle 7/64$, but I don't know how to get it.

2. ## Re: Conditional probability + Poisson question

You need to find $\displaystyle P(X=6\mid Z=8)$, which is equal to $\displaystyle \frac{P(X=8\cap Z=8)}{P(Z=8)}$. The event $\displaystyle X=6\cap Z=8$ is the same as $\displaystyle X=6\cap Y=2$, and, since $\displaystyle X$ and $\displaystyle Y$ are independent, $\displaystyle P(X=6\cap Y=2)=P(X=6)P(Y=2)$.

3. ## Re: Conditional probability + Poisson question

Originally Posted by emakarov
You need to find $\displaystyle P(X=6\mid Z=8)$, which is equal to $\displaystyle \frac{P(X=8\cap Z=8)}{P(Z=8)}$. The event $\displaystyle X=6\cap Z=8$ is the same as $\displaystyle X=6\cap Y=2$, and, since $\displaystyle X$ and $\displaystyle Y$ are independent, $\displaystyle P(X=6\cap Y=2)=P(X=6)P(Y=2)$.
No, X and Y are not independent, since we know X+Y = 8.

Given that X+Y=8, X has a Binomial distribution with p = 1/2 and n = 8.

4. ## Re: Conditional probability + Poisson question

Originally Posted by awkward
No, X and Y are not independent, since we know X+Y = 8.
I am saying that $\displaystyle P(X=6\cap Y=2)=P(X=6)P(Y=2)$, not that $\displaystyle P(X=6\cap Y=2\mid X+Y=8)$$\displaystyle =P(X=6\mid X+Y=8)P(Y=2\mid X+Y=8)$.

5. ## Re: Conditional probability + Poisson question

OK, at first I didn't see your point, but now I see you are correct.