Originally Posted by

**Moo** Hello,

Why not ?

You can compute E[X] and E[Y] I guess ^^ (same binomial distributions)

$\displaystyle \mathbb{E}(XY)=\sum_{k=0}^n\sum_{l=0}^n k \cdot l \cdot \mathbb{P}(X=k,Y=l)=\sum_{k=0}^n\sum_{l=0}^n k \cdot l \cdot \mathbb{P}(Y=l|X=k)\mathbb{P}(X=k)$

$\displaystyle =\sum_{k=0}^n k \cdot \mathbb{P}(X=k) \sum_{l=0}^n l \cdot \mathbb{P}(Y=l|X=k)$

For the distribution $\displaystyle \mathbb{P}(Y=l|X=k)$, note that it's equivalent to saying "what is the probability to get l 2's, while there already had k 1's ?

You can also see that l can get values only between 0 and n-k.

And that it follows a binomial distribution (n-k,1/5) ~ Z

So we actually have :

$\displaystyle \mathbb{E}(XY)=\sum_{k=0}^n k \cdot \mathbb{P}(X=k) \sum_{l=0}^{n-k} l \cdot \mathbb{P}(Z=l)$

But $\displaystyle \sum_{l=0}^{n-k} l \cdot \mathbb{P}(Z=l)$ looks like a lot $\displaystyle \mathbb{E}(Z)$, doesn't it ? :D

Can you try to do this ?