1. ## Re: Combination or Permutation

Originally Posted by BGM
Does this formula consider both variables already (Ie., the probability of X and Y together)?

2. ## Re: Combination or Permutation

I guided myself with this one http://bit.ly/vtozxA. But that doesn't guide me how to get the probabilities of BOTH X and Y occurring. I have to compute separately for X and Y using the given formula in that site. And to get their joints, I had to multiply the separate probabilities computed which you said is not appropriate. Please tell me how do we get the probabilities when both occur together (ie. X=2 and Y=1; Y=2 and X=1). Getting P(X and Y) is a joint probability and that's what I need. The correlation is given in this problem and it is -1. When I tried to compute for the correlation, it does not equal to -1 so I concluded that indeed what I did was wrong.

3. ## Re: Combination or Permutation

First, you should know that if and only if .

So both events are equivalent, and therefore

The formula I given to you is usually known as the probability mass function of an univariate hypergeometric distribution (that is the usual notion).
Please do not think it is a two dimensional problem - it is a univariate problem as I have emphasized several times already.

So anyway you should know that the support of is and evaluate the pmf at these points

4. ## The Following User Says Thank You to BGM For This Useful Post:

dEconomist (12-31-2011)

5. ## Re: Combination or Permutation

Your formula is different from my readings. Yes, I know that X+Y=3. Why there are two rows inside the parentheses? What is the operation to be used for that? Is that equivalent to (kCx)(N-kCn-x) all over by NCn? The variables before and after C here are subscripts.

6. ## Re: Combination or Permutation

And I know a bit of multinomial but isn't that for with repetition only?

7. ## Re: Combination or Permutation

I think I get it. The probabilities obtained using the hypergeometric distribution is good as the joint probability already. Say, we get P(X=0), using hypergeometric, it is 5/12, then that means we actually obtain 3 of Ys. And when P(Y=2), that is 1/2, then that means there is one of X.

8. ## Re: Combination or Permutation

Yes I think you have got it.