I am working on a problem and I have no clue where to start.
I'm not sure what It is asking, or where to start.
If you guys could give me the steps to take, show me what concepts are used, or a link to somewhere helpful I would appreciate it!
I am lost so any help is appreciated!
That table is a joint probability distribution table for $X$ and $Y$. For example, the first empty box has $P(X=1 \cap Y=1)$.
I'll give two methods to calculate all the joint probabilities. The first a more obvious method and the second perhaps a more intuitive method.
Solution $1$
We mostly use the formula definition for conditional probability in various ways, which is: for events $A,B$, if $P(B) \neq 0$,
$$P(A \mid B) = \dfrac{P(A\cap B)}{P(B)}.$$
Firstly,
\begin{eqnarray*} P(Y=1) &=& \sum_{k=1}^3{P(Y=1\vert X=k)P(X=k)} \qquad\qquad\text{(by Law of Total Probability)} \\ && \\ &=& \dfrac{3}{5}\left(P(X=1) + P(X=2) + P(X=3)\right) = \dfrac{3}{5}. \end{eqnarray*}
We can use that to get three of the six required values, \begin{eqnarray*} P(X=1 \cap Y=1) &=& P(X=1\mid Y=1)P(Y=1) = \frac{1}{6}\cdot\frac{3}{5} = \frac{1}{10}. \\ &&\\ P(X=2 \cap Y=1) &=& P(X=2\mid Y=1)P(Y=1) = \frac{2}{6}\cdot\frac{3}{5} = \frac{2}{10}. \\ &&\\ P(X=3 \cap Y=1) &=& P(X=3\mid Y=1)P(Y=1) = \frac{3}{6}\cdot\frac{3}{5} = \frac{3}{10}. \\ \end{eqnarray*}
Next, \begin{eqnarray*} P(X=1) &=& \dfrac{P(X=1\cap Y=1)}{P(Y=1\mid X=1)} = \dfrac{1/10}{3/5} = \frac{1}{6}. \\ &&\\ P(X=2) &=& \dfrac{P(X=2\cap Y=1)}{P(Y=1\mid X=2)} = \dfrac{2/10}{3/5} = \frac{2}{6}. \\ &&\\ P(X=3) &=& \dfrac{P(X=3\cap Y=1)}{P(Y=1\mid X=3)} = \dfrac{3/10}{3/5} = \frac{3}{6}. \\ \end{eqnarray*}
\begin{eqnarray*} P(Y=2\mid X=1) &=& 1-P(Y=1\text{ or }3\mid X=1) = 1-\frac{3}{5} = \frac{2}{5}. \\ &&\\ P(Y=3\mid X=2) &=& 1-P(Y=1\text{ or }2\mid X=2) = 1-\frac{3}{5} = \frac{2}{5}. \\ &&\\ P(Y=2\mid X=3) &=& 1-P(Y=1\text{ or }3\mid X=3) = 1-\frac{3}{5} = \frac{2}{5}. \\ \end{eqnarray*}
We can now calculate the last three required probabilities. \begin{eqnarray*} P(X=1\cap Y=2) &=& P(Y=2\mid X=1)P(X=1) = \frac{2}{5}\cdot\frac{1}{6} = \frac{1}{15}. \\ &&\\ P(X=2\cap Y=3) &=& P(Y=3\mid X=2)P(X=2) = \frac{2}{5}\cdot\frac{2}{6} = \frac{2}{15}. \\ &&\\ P(X=3\cap Y=2) &=& P(Y=2\mid X=3)P(X=3) = \frac{2}{5}\cdot\frac{3}{6} = \frac{3}{15}. \\ &&\\ \end{eqnarray*}
So we get: $$ \begin{array}{c|ccc} Y & X=1 & X=2 & X=3 \\ \hline 1 & 1/10 & 2/10 & 3/10 \\ 2 & 1/15 & 0 & 3/15 \\ 3 & 0 & 2/15 & 0 \end{array} $$
Solution $2$
Note that
$$\dfrac{P(X=i \mid Y=k)}{P(X=j \mid Y=k)} = \dfrac{P(X=i \cap Y=k)P(Y=k)}{P(X=j \cap Y=k)P(Y=k)} = \dfrac{P(X=i \cap Y=k)}{P(X=j \cap Y=k)}.$$
We see that the ratio of conditional probabilities equals the ratio of the corresponding joint probabilities.
In the first column we see that, $$P(Y=2\mid X=1) = 1-P(Y=1\text{ or }3\mid X=1) = 1-\frac{3}{5} = \frac{2}{5}.$$ Thus,
\begin{eqnarray*} P(Y=1\cap X=1) : P(Y=2\cap X=1) &=& P(Y=1\mid X=1) : P(Y=2\mid X=1) \\ &=& \frac{3}{5} : \frac{2}{5} = 3 : 2. \end{eqnarray*}
So, for some $r \gt 0$, we must have $$ \begin{array}{c|ccc} Y & X=1 & X=2 & X=3 \\ \hline 1 & 3r & ? & ? \\ 2 & 2r & 0 & ? \\ 3 & 0 & ? & 0 \end{array} $$
Working now in the $Y=1$ row,
\begin{eqnarray*} && P(X=1\cap Y=1) : P(X=2\cap Y=1) : P(X=3\cap Y=1) \\ &&\qquad \qquad \qquad = P(X=1\mid Y=1) : P(X=2\mid Y=1) : P(X=3\mid Y=1) \\ &&\qquad \qquad \qquad = \frac{1}{6} : \frac{2}{6} : \frac{3}{6} \\ &&\qquad \qquad \qquad = 1:2:3. \\ \end{eqnarray*}
So now we have, $$ \begin{array}{c|ccc} Y & X=1 & X=2 & X=3 \\ \hline 1 & 3r & 6r & 9r \\ 2 & 2r & 0 & ? \\ 3 & 0 & ? & 0 \end{array} $$
Working in the remaining columns as we did in the $X=1$ column, we can fill in the last two ? boxes: $$ \begin{array}{c|ccc} Y & X=1 & X=2 & X=3 \\ \hline 1 & 3r & 6r & 9r \\ 2 & 2r & 0 & 6r \\ 3 & 0 & 4r & 0 \end{array} $$
We know all the probabilities in the table must add to $1$, so we have $30r=1$ and so $r=1/30$. Finally, we can complete the table (the same as Solution $1$): $$ \begin{array}{c|ccc} Y & X=1 & X=2 & X=3 \\ \hline 1 & 1/10 & 2/10 & 3/10 \\ 2 & 1/15 & 0 & 3/15 \\ 3 & 0 & 2/15 & 0 \end{array} $$