# Joint probability function (3 Cards without replacement)

I have the following problem:

3 Cards are picked from a deck without replacement. Let $X$ number of hearts and $Y$ number of diamonds. Find the joint probability function.

I know $X$ and $Y$ are both discrete random variables that can take on 0, 1, 2 and 3. Therefore:

$$\mathbb{P}\left(X=i,Y=j\right)=??\qquad\forall i,j\in\left\{ 0,1,2,3\right\}$$

You seek the probability for selecting $i$ from $13$ hearts, $j$ from $13$ diamonds, and $3-i-j$ from $26$ other suits when selecting $3$ from $52$ cards.
• $$\mathbb{P}\left(X=i,Y=j\right)=\dfrac{\left(\begin{array}{c} 13\\ i \end{array}\right)\left(\begin{array}{c} 13\\ j \end{array}\right)\left(\begin{array}{c} 26\\ 2-i-j \end{array}\right)}{\left(\begin{array}{c} 52\\ 2 \end{array}\right)}\qquad\forall i,j\in\left\{ 0,1,2,3\right\}$$ I'm not sure. It's ok? Feb 13 '18 at 23:41
• Sorry. Now??$$\mathbb{P}\left(X=i,Y=j\right)=\dfrac{\left(\begin{array}{c} 13\\ i \end{array}\right)\left(\begin{array}{c} 13\\ j \end{array}\right)\left(\begin{array}{c} 26\\ 3-i-j \end{array}\right)}{\left(\begin{array}{c} 52\\ 3 \end{array}\right)}\qquad\forall i,j\in\left\{ 0,1,2,3\right\}$$ Feb 13 '18 at 23:52