<s>[INST] Finding conditional probability from the joint PMF I have no idea how to get tabular to work on here so the table isn't rendering.

A joint PMF $p_{X, Y}[i, j]$ has the values shown in table. Determine the conditional PMF $p_{Y|X}$. Are the random variables independent?
$$
\begin{array}{|c|c|c|c|}
  \hline
  & j = 1 & j = 2 & j = 3\\
  \hline
  i = 1 & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{2}{10}\\
  \hline
  i = 2 & \tfrac{1}{20} & \tfrac{1}{20} & \tfrac{1}{10}\\
  \hline
  i = 3 & \tfrac{3}{10} & \tfrac{1}{20} & \tfrac{1}{20}\\
  \hline
\end{array}
$$

When I follow the formula, the conditional probabilities aren't summing to one.

The $p_{Y|X = x_i}[j|x_i] = \frac{p_{X, Y}[x_i, j]}{p_X[x_i]}$ for all $j$ and similarly for $p_{X|Y = y_j}$.
  \begin{alignat*}{4}
    p_{Y|X = 1}[j|1] &=
    \begin{cases}
      \frac{3}{10}, & j = 1\\
      \frac{3}{10}, & j = 2\\
      \frac{6}{10}, & j = 3
    \end{cases} & \qquad &
    p_{Y|X = 2}[j|2] &&={}
    \begin{cases}
      \frac{3}{20}, & j = 1\\
      \frac{3}{20}, & j = 2\\
      \frac{3}{10}, & j = 3
    \end{cases}\\
    p_{Y|X = 3}[j|3] &=
    \begin{cases}
      \frac{9}{10}, & j = 1\\
      \frac{3}{20}, & j = 2\\
      \frac{3}{20}, & j = 3
    \end{cases} & \qquad &
    p_{X|Y = 1}[i|1] &&={}
    \begin{cases}
      \frac{3}{10}, & i = 1\\
      \frac{3}{20}, & i = 2\\
      \frac{9}{10}, & i = 3
    \end{cases}\\
    p_{X|Y = 2}[i|2] &=
    \begin{cases}
      \frac{3}{10}, & i = 1\\
      \frac{3}{20}, & i = 2\\
      \frac{3}{20}, & i = 3
    \end{cases} & \qquad &
    p_{X|Y = 3}[i|3] &&={}
    \begin{cases}
      \frac{6}{10}, & i = 1\\
      \frac{3}{10}, & i = 2\\
      \frac{3}{20}, & i = 3
    \end{cases}
  \end{alignat*}
 [/INST] I'm not sure where you are getting those figures.   The marginal probability of $X$ is the sum of the joint probabilities for all values of $Y$ at particular values of $X$.
$\begin{align}
\text{Thus:}
\\ \because p_X(i) & = \sum_{j=1}^3 p_{X,Y}(i,j)
\\ &= p_{X,Y}(i,1)+p_{X,Y}(i,2)+p_{X,Y}(i,3) 
\\[1ex]
\therefore p_X(1) &= p_{X,Y}(1,1)+p_{X,Y}(1,2)+p_{X,Y}(1,3) 
\\ &= \tfrac 1{10}+\tfrac 1{10}+\tfrac 2{10}
\\ &= \tfrac 2 5
\\ \text{Similarly:}
\\[2ex]
\because p_{Y\mid X}(j\mid i)&= \frac{p_{X,Y}(i, j)}{p_X(i)}
\\[2ex]
\therefore p_{Y\mid X}(j\mid 1)&= \frac{p_{X,Y}(1, j)}{p_X(1)}
\\ & = \frac{5}{2}\begin{cases}\frac 1{10} & :j=1\\\frac 1{10} &:j=2\\\frac 2{10} &:j=3\end{cases}
\\ & = \begin{cases}\frac 1{4} & :j=1\\\frac 1{4} &:j=2\\\frac 1{2} &:j=3\end{cases}
\end{align}$
You can do the rest now, right?
</s>