Your evil probability professor has an urn with 9 balls. Your evil probability professor has an urn with 9 balls:  2 red, 3 white and 4 blue.  He draws two balls from the urn without replacement.  Let X be the number of red balls drawn and Y the number of white balls.
a) Determine the joint probability mass function of X and Y. 
b) Are X and Y independent random variables? 
c) Compute the covariance between X and Y.  
For point A:  
$P(0,0)=\frac{4}{9} \cdot \frac{3}{8} = \frac{1}{6}$ That is correct for the solution. 
$P(0,1)=\frac{4}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{4}{8} = \frac{1}{3}$ That is correct for the solution. 
$P(1,0)=\frac{2}{9} \cdot \frac{4}{8} + \frac{4}{9} \cdot \frac{2}{8}= \frac{2}{9}$ That is correct for the solution.  
$P(1,1)=\frac{2}{9} \cdot \frac{3}{8} +  \frac{3}{9} \cdot \frac{2}{8}= \frac{1}{6}$  That is correct for the solution. 
$P(2,0)=\frac{2}{9} \cdot \frac{1}{8} = \frac{1}{36}$ That is correct for the solution. 
$P(0,2)=\frac{3}{9} \cdot \frac{2}{8} = \frac{1}{12}$ That is correct for the solution. 
For point B to check the independancy I have just to check if for example $P(X=0,Y=0) = P(X=0) \cdot P(Y=0)$. 
$\frac{1}{6} \neq (\frac{7}{9} \cdot \frac{6}{8}) \cdot (\frac{6}{9} \cdot \frac{5}{8})$. 
So X and Y are not independent. 
For point C I know that the covariance $Cov(X,Y)=E[X \cdot Y]-E[X]\cdot E[Y]$, but how can compute the expectations, do I have to figure put with distribution is? How can I do it?
Can someone help me?
Thanks in advance, Fabio!
 A: For a) keep in mind that the order you draw the balls in matters. For example, for $P(0,1)$ we can draw a blue ball then a white ball or a white ball then a blue ball.
For b) independence says that for every $x$ and $y$, $P(x,y) = P_X(x)P_Y(y)$. It's not enough to check a single pair $(x,y)$ to prove independence. On the other hand, if there is some $x$ and $y$ for which $P(x,y) \ne P_X(x)P_Y(y)$ then $X$ and $Y$ are not independent.
For c) recall that $\displaystyle \mathbf E[f(X,Y)] = \sum_{(x,y)} f(x,y)P(x,y)$.
A: Another way to check dependence (i.e. part B only) is that if $X,Y$ are independent then for any values $x,y$ we have $P(Y=y) = P(Y=y | X=x)$.  However, clearly if there are 2 red balls then there cannot be any white balls, i.e. $P(Y > 0) > 0$ but $P(Y > 0 | X=2) = 0$.  This might be a more intuitive example demonstrating dependence.
A: 
but how can compute the expectations, do I have to figure put with distribution is?

You have the distribution, $P(x,y)$.  Use it.
The expectation of function $g$ of $X,Y$ is:
$$\begin{align}\mathsf E(g(X,Y)) &=\sum_{x}\sum_{y} g(x,y)~P(x,y)\\[1ex]\textsf{so...}\\[2ex]\mathsf E(X) &=\sum_{x}\sum_{y} x~P(x,y) \\ &=0+P(1,0)+P(1,1)+2P(2,0)\end{align}$$
And so on.
