An urn contains $a$ white, $b$ black and $c$ red balls.

Picking $n$ balls randomly and without replacement.

$X,Y,Z$ denote the number of the white, black and red balls.


1) joint distribution of $X,Y$ and $Z$

2) marginal distribution ( distribution of $ X$ )

3) $\mathbb{E}(XY)$

My thoughts:

So I think, that we consider a hypergeometric distribution because we picking balls randomly without replacement. So maybe 1) is $P(X=s,Y=t,Z=u) = \frac{\binom{a}{s} \binom{b}{t} \binom{c}{u}}{\binom{a+b+c}{n}} $ with $ s+t+u=n $.

2) has to be $ P(X = s)= \frac{\binom{a}{s} \binom{b+c-a}{n-s}}{\binom{a+b+c}{n}} $.

3) $ \mathbb{E}(XY) $ = $ \sum_{(x,y)} xy P(X=x,Y=y).$ Maybe we can replace now P(X=x,Y=y).I'm not sure. I'm stucked here.

Am I right with 1) and 2) ? How can I "prove" it?

  • $\begingroup$ I do not think you have to "prove" it, just write down the (joint) pmf would be fine. Specify clearly about the support, and then 1) looks good. For 2) you do not need to $-a$. For 3), the common techniques will be decompose $X, Y$ into sum of $n$ (dependent) indicators respectively and calculate. $\endgroup$ – BGM Dec 5 '17 at 9:03
  • $\begingroup$ 1) and 2) is clear now. :) Thanks. Could you give me only the beginning of 3) ? So I can try the rest on my own. $\endgroup$ – Mugumble Dec 5 '17 at 9:54

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