Finding the joint Probability distribution of $X$ and $Y$? If the joint probability distribution of X and Y is given by
$$f(x,y)= \frac{(x-y)^2}{7}, \text{for }x=1,2,3;y=1,2 $$
$(1)$ Find the probability distribution of $U = X + Y; $
$(2)$ Find the conditional probability distribution of $X$ given $U =4.$
In order to solve this problem one must draw a chart.
$$\begin{array}{|c|c|c|c|}
\hline
(x,y)& (1,1) & (1,2) & (2,1) & (2,2) & (3,1) & (3,2) \\ \hline
 f(x,y)& 0 & \frac{1}{7} & \frac{1}{7} & 0 & \frac{4}{7} & \frac{1}{7}\\ \hline
  U=x+y& 2 & 3 &3 & 4 & 4 & 5\\ \hline
x &  & &\\ \hline
\end {array}$$
How does one fill up the rest of the table and answer questions one and two.
EDIT
In order to find solve $(1)$ one must add all the related $f(x,y)$ relations. Thus
$(1)$ $$ \quad P(U=2) =0, \\ P(U=3) = \frac{1}{7} + \frac{1}{7} = \frac27, \\ P(U=4)= 0+\frac{4}{7} = \frac47, \\ P(U=5)=\frac{1}{7}$$
One must use this notation to solve.
$$P(x=1|U=4)= \frac{P(x=1,U=4)}{P(U=4)} = ? \\P(x=2|U=4) = ? \\ P(x=3|U=4) = ? $$
Knowing this does anyone know how to solve $(2)$ using this notation? What does one substitute for this question to derive the answer?
 A: Ok so the first thing you notice is that so far your attempt has
$$
\begin{align}
P(U = 2) &= 0, \\
P(U = 3) &= \frac{1}{7}, \\
P(U = 4) &= \frac{4}{7}, \\
P(U = 5) &= \frac{1}{7}
\end{align}
$$
and zero elsewhere, but summing over all possible situations only takes us to $\frac{6}{7}$ so something has clearly gone wrong! So what you have missed is that
$$
P(U=3) = P(x=1,y=2) + P(x=2,y=1) = \frac{2}{7}.
$$
For the second part of your question look at your table and study the different combinations of $x,y$ that will make $U=4$ and then look at the joint probability of these combinations, and you should see clearly what the distribution of $x$ must be.
A: Here is a completed table of the discrete probabilities:
$$ \begin{array}{|c|c|c|c|}
\hline
(x,y)& (1,1) & (1,2) & (2,1) & (2,2) & (3,1) & (3,2) \\ \hline
 f(x,y)& 0 & \frac{1}{7} & \frac{1}{7} & 0 & \frac{4}{7} & \frac{1}{7}\\ \hline
  U=x+y& 2 & 3 &3 & 4 & 4 & 5\\ \hline
x & 1 & 1 & 2 & 2 & 3 & 3 \\ \hline
\end {array} $$
The Reader will recognize that the last row's entries are simply the $x$ values of the coordinate pairs in the first row.  The intent of forming such a row is likely to help answer the second part of the exercise given in the Question, what is the conditional probability distribution of $X$, given that the sum $U = 4$?
The table makes it clear that while the events $X=2,Y=2$ and $X=3,Y=1$ both lead to $U=4$, the former of these events is assigned a zero probability (and the latter is assigned a positive probability).  Therefore when the conditional probabilities are assigned, conditioned on $U=4$, we have:
$$ \mathbf{Pr}(X=2 \mid U=4) = 0 \; ; \; \mathbf{Pr}(X=3 \mid U=4) = 1 $$
