Two hats are drawn randomly w/o replacement from box containing $8$ black, $4$ red, and $2$ yellow hats. If $X$ denotes the number of black hats drawn and $Y$ the number of red hats drawn. What is the joint probability function $f(x,y)$ of $X$ and $Y$? So I have to give the values of $X$ and $Y$ and their probabilities.
Here's my thoughts: $$\begin{align} X & = {{8 \choose x}{6 \choose 2-x}}\bigg/{{14 \choose 2}} \\[1ex] Y & = {{4 \choose x}{10 \choose 2-x}}\bigg/{{14 \choose 2}} \end{align}$$
Is that the values of $X$ and $Y$? Not sure what the exactly the proper answer is for the question. Thanks in advance.
$\checkmark$ so far. What you have are actually the probability measures of $X$ and $Y$ at certain integer values.
For $X$: you are selecting $x\in\{0,1,2\}$ of 8 black hats out of all the ways to select 2 hats from 14. Likewise for $Y$: you are selecting $y\in\{0,1,2\}$ of 4 red hats out of the same probability space.
$$\begin{align} f_X(x)=\mathsf P(X=x) & = {{8 \choose x}{6 \choose 2-x}}\bigg/{{14 \choose 2}} & \text{if } x\in \{0,1,2\} \\[1ex] f_Y(y)=\mathsf P(Y=y) & = {{4 \choose y}{10 \choose 2-y}}\bigg/{{14 \choose 2}} & \text{if } y\in \{0,1,2\} \end{align}$$
However you require the joint probability measure. $f(x,y)=\mathsf P(X=x, Y=y)$ (Note: these are not independent random variables, so you cannot use the product rule.)
You need to count ways to select both $x\in\{0,1,2\}$ of 8 black hats AND $y\in\{0,1,2\}$ of 4 red hats out of the same probability space. There is an additional constraint that $x+y\leq 2$.
$$\begin{align} f(x,y)=\mathsf P(X=x,Y=y) & = \boxed{?} & \text{if } x\in \{0,1,2\}, y\in\{0\ldots 2-x\} \end{align}$$
Can you fill in the box now?