<s>[INST] How to calculate joint probability distribution for replacement sample? If 3 cards drawn with replacement from 12 face card and $X$ represents the no. of kings`
and $Y$ represents no of jack construct joint probability distribution.
What I am doing;
12 Face cards having $4K$, $4Q$ and $4J$
$$p(K) = \frac{4}{12} \\    
p(Q) = \frac{4}{12}   \\
p(J) = \frac{4}{12}$$
Where i am not getting clue to solve further help required?
For $x=0$ and $y =0$  the sampling will be done from queen and it is with replacement so it may be this?  $$\left(\frac{4}{12}\right)^3$$
 [/INST] Record the results in order. For example, KJJ means we got a King, then a Jack, then a Jack. There are $3^3$ such sequences, all equally likely.
Now for all possible values of $x$ and $y$, we find the number of ways to have $x$ Kings and $y$ Jacks. We can make a list. It should be systematic, so we do not leave out any cases.  Or else we can use formulas. I think at this stage a list is better, more concrete. But it is lengthy. We can save time by taking advantage of symmetry. It is enough to find the probabilities when $x\le y$, since the probability of $a$ Kings and $b$ Jacks is the same as the probability of $b$ Kings and $a$ Jacks. 
(i) $x=0$, $y=0$. There is $1$ way to have $0$ K and $0$ J. The probability is $\frac{1}{3^3}$.
(ii) $x=0$, $y=1$. There are $3$ ways to have $0$ K and $1$ J, for the J can be put in any of $3$ places. The probability is $\frac{3}{3^3}$.
For free, we get that the probability that $x=1$ and $y=0$ is $\frac{3}{3^3}$.
(iii) $x=0$, $y=2$. There are $3$ ways to place the two J. For free, we get the answer for $x=2$, $y=0$
(iv) $x=0$, $y=3$. There is only $1$ way to get this. For free, we get the answer for $x=3$, $y=0$. 
(v) $x=1$, $y=1$. The K can be placed in $3$ ways, and for each such way there are $2$ ways to place the J, for a total of $6$. Thus the probability is $\frac{(3)(20}{3^3}$.  
(vi) $x=1$, $y=2$. The King can be placed in $3$ ways. So the probability is $\frac{3}{27}$. For free we get the answer for $x=2$, $y=1$. 
To see that we got them all, it is good to insert the numbers in a $3\times 3$ table, and check that our computed probabilities add up to $1$. 
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