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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 $cov[X,Y]$

$E[XY]=(1)(2)(\frac{1}{6})+(1)(3)(\frac{1}{6})+(2)(2)(\frac{1}{6})+(2)(3)(\frac{1}{6})+(3)(3)(\frac{2}{6})=\frac{33}{6}$

$E[X]=(1)(\frac{2}{6})+(2)(\frac{2}{6})+(3)(\frac{2}{6})=2$

$E[Y]=(2)(\frac{2}{6})+(3)(\frac{4}{6})=\frac{16}{6}$

$cov[X,Y]=\frac{33}{6}-(2)(\frac{16}{6})=\frac{1}{6}$

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    $\begingroup$ There's just a small typo in the denominator in $(b)$. $\endgroup$ Aug 29, 2013 at 11:57

1 Answer 1

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The solution is correct. Congratulations.

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