<s>[INST] Let $X$ be the number on the first ball drawn and $Y$ the larger of the two numbers draw. 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 number on the first ball drawn and $Y$ the larger of the two numbers draw
(a) Find the joint discrete density function of $X$ and $Y$.
$$\begin{array}{c|cc|c}
x/y & 2 & 3 & f_x(x)\\
\hline
\\ 1 & 1/6 & 1/6 & 2/6
\\ 2 & 1/6 & 1/6 & 2/6
\\ 3 & 0 & 2/6 & 2/6
\\
\hline
\\ f_y(y) & 2/6 & 4/6 &
\end{array}
$$
(b) Find $P[X=1|Y=3]$
$$P(X=1|Y=3)=\frac{P(X=1,Y=3)}{P(Y=3)}=\frac{1}{4}$$
(c) Find $cov[X,Y]$
$E[XY]=(1)(2)(\frac{1}{6})+(1)(3)(\frac{1}{6})+(2)(2)(\frac{1}{6})+(2)(3)(\frac{1}{6})+(3)(3)(\frac{2}{6})=\frac{33}{6}$
$E[X]=(1)(\frac{2}{6})+(2)(\frac{2}{6})+(3)(\frac{2}{6})=2$
$E[Y]=(2)(\frac{2}{6})+(3)(\frac{4}{6})=\frac{16}{6}$
$cov[X,Y]=\frac{33}{6}-(2)(\frac{16}{6})=\frac{1}{6}$
 [/INST] The solution is correct. Congratulations.
</s>