Distribution of $Y$ given $X$ Let $X_1$ and $X_2$ be the numbers on two independent fair-die rolls. Let $X$ be the minimum and $Y$ the maximum of $X_1$ and $X_2$.
$(a)$ Find the distribution of $Y\mid X$
Here is my work:
Here is what I know:
$$P(Y = y\mid X=x) = \frac{P(Y=y, X=x)}{P(X=x)}$$
and $P(X=x)$ can be found by the following equation $\sum_y p_{X,Y}(x,y)$
So here is what I did:
\begin{array}{cc|cccc}
&&&&&y \\
&& 1 & 2 & 3 & 4 & 5 & 6 & P(X=x)\\
\hline
&1 & \frac{1}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{11}{36} \\
&2 & 0 & \frac{1}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{9}{36} \\
x&3 & 0 & 0 & \frac{1}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{7}{36} \\
&4 & 0 & 0 & 0 & \frac{1}{36} & \frac{2}{36} & \frac{2}{36} & \frac{5}{36} \\
&5 & 0 & 0 & 0 & 0 & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} \\
&6 & 0 & 0 & 0 & 0 & 0 & \frac{1}{36} & \frac{1}{36}
\end{array}
So I said:
$$
P(Y=y\mid X=x) = 
    \begin{cases}
       \frac{1}{36} \div P(X=x) & \text{for } x = y \\[4pt]
       \frac{2}{36} \div P(X=x) & \text{for } x = 1,2,3,4,5,6, \ y > x
    \end{cases}
$$
Am I correct?


So I said, $$P(Y=y\mid X=x) = 
    \begin{cases}
       \frac{1}{36} \cdot P(X=x) & \text{for } x = y \\[4pt]
       \frac{2}{36} \cdot P(X=x) & \text{for } x = 1,2,3,4,5,6, \ y > x
    \end{cases}$$
  Am I correct? 

Almost but, no.   You know that the conditional probability is the joint divided by the marginal, so why did you multiply rather than divide?
Also notice that: $\mathsf P(X=x) = \frac{13-2x}{36}$
$$\begin{align}\mathsf P(Y{=}y\mid X{=}x) & = \begin{cases} \frac{1}{36} \div P(X=x) & \text{for } x\in\{1\,\ldots\,6\}{\times}\{x\}\\[1ex] \frac{2}{36} \div P(X=x) & \text{for } (x,y)\in \{1\,\ldots\,5\}{\times}\{x{+}1\,\ldots\,6\}  \end{cases} \\[2ex] & = \begin{cases}\frac{1}{13-2x} & \text{for } (x,y)\in\{1\,\ldots\,6\}{\times}\{x\}\\[1ex] \frac{2}{13-2x} & \text{for } (x,y)\in \{1\,\ldots\,5\}{\times}\{x{+}1\,\ldots\,6\}  \end{cases}\end{align}$$
