2 Cards are picked from a deck without replacement. Let X= number of aces, and Y= number of kings. Find the joint probability function (in a 3×3 table)
X and Y are both discrete random variables that can take on 0,1 and 2.
There are 5 choose 2 (1326) ways to pick 2 cards. The probability of getting an ace are $\frac{4}{52}$ (also the probability of getting a king).
How can I use this to find the joint probability function in a table? What kind of probability distribution can I use?
Edit: would the technique used here be similar to the question answered here: 5 cards / Joint Probability Function
I've also filled out three blocks in the table, since you can't have 2 aces and 2 kings, or 2 kings 1 aces, or 2 aces 1 king.
Best Answer
Split up like $52=4+4+44$, i.e. $4$ kings $4$ aces and the other $44$ cards.
In total $N:=\binom{52}2$ different pair-picks are possible and equiprobable.
You can find the probabilities by dividing the outcomes by $N$.
edit (alternative):
Pick two cards one by one.
For $i=1,2$ let $A_{i},K_{i},E_{i}$ denote the events that the $i$-pick gives ace, king and other card respectively.
$P\left(A_{1}\cap A_{2}\right)=P\left(A_{1}\right)P\left(A_{2}\mid A_{1}\right)=\frac{4}{52}\frac{3}{51}$
$P\left(A_{1}\cap K_{2}\right)=P\left(A_{1}\right)P\left(K_{2}\mid A_{1}\right)=\frac{4}{52}\frac{4}{51}$
$P\left(A_{1}\cap E_{2}\right)=P\left(A_{1}\right)P\left(E_{2}\mid A_{1}\right)=\frac{4}{52}\frac{44}{51}$
$P\left(K_{1}\cap A_{2}\right)=P\left(K_{1}\right)P\left(A_{2}\mid K_{1}\right)=\frac{4}{52}\frac{4}{51}$
$P\left(K_{1}\cap K_{2}\right)=P\left(K_{1}\right)P\left(K_{2}\mid K_{1}\right)=\frac{4}{52}\frac{3}{51}$
$P\left(K_{1}\cap E_{2}\right)=P\left(K_{1}\right)P\left(E_{2}\mid K_{1}\right)=\frac{4}{52}\frac{44}{51}$
$P\left(E_{1}\cap A_{2}\right)=P\left(E_{1}\right)P\left(A_{2}\mid E_{1}\right)=\frac{44}{52}\frac{4}{51}$
$P\left(E_{1}\cap K_{2}\right)=P\left(E_{1}\right)P\left(K_{2}\mid E_{1}\right)=\frac{44}{52}\frac{4}{51}$
$P\left(E_{1}\cap E_{2}\right)=P\left(E_{1}\right)P\left(E_{2}\mid E_{1}\right)=\frac{44}{52}\frac{43}{51}$
As an example note that e.g. $$P(A=1,K=1)=P(A_1\cap K_2)+P(K_1\cap A_2)=\frac{4}{52}\frac{4}{51}+\frac{4}{52}\frac{4}{51}$$