Let $X$ be the smaller of the two numbers drawn and $Y$ the larger. Find the joint discrete density function of $X$ and $Y$ Consider a sample of size 2 drawn without replacement from an urn containing three balls, numbered 1,2, and 3. Let $X$ be the smaller of the two numbers drawn and $Y$ the larger.
(a) Find the joint discrete density function of $X$ and $Y$. $$\begin{array}{c|cc|c} x/y & 2 & 3 & f_x(x)\\ \hline \\ 1 & 2/6 & 2/6 & 4/6 \\ 2 & 0 & 2/6 & 2/6 \\ \hline \\ f_y(y) & 2/6 & 4/6 & \end{array} $$ 
(b) Find the conditional distribution of $Y$ given $X=1$ 
$$P(Y=2|X=1)=\frac{P(X=1,Y=2)}{P(X=1)}=\frac{1}{2}$$
$$P(Y=3|X=1)=\frac{P(X=1,Y=3)}{P(X=1)}=\frac{1}{2}$$
(c) Find $cov[X,Y]$$
$E[XY]=(1)(2)(\frac{2}{6})+(1)(3)(\frac{2}{6})+(2)(3)(\frac{2}{6})=\frac{22}{6}$
$E[X]=(1)(\frac{4}{6})+(2)(\frac{2}{6})=\frac{8}{6}$ 
$E[Y]=(2)(\frac{2}{6})+(3)(\frac{4}{6})=\frac{16}{6}$
$$cov[X,Y]=\frac{22}{6}-(\frac{8}{6})(\frac{16}{6})=\frac{1}{9}$$
Could someone tell me if my answers are corrects? Thanks for your help :) 

Almost everything is right. There is only an error of arithmetic in calculating the covariance.
I do not know why you consistently used $\frac{2}{6}$ instead of $\frac{1}{3}$.
