What does is mean $Y=\min \{X_1,X_2\}$? $Y=\min \{X_1,X_2\}$ and $Z=\max \{X_1,X_2\}$?
Let's determine $X_1$ distribution to be $\operatorname{Bin}(2,\frac{1}{2})$ and the distribution of $X_2$ to be $U(1,2,3)$. Also $X_,X_2$ are independent. 
After some calculations:


*

*$P(X_1=0)=\frac{1}{4}$

*$P(X_1=1)=\frac{1}{2}$

*$P(X_1=2)=\frac{1}{4}$


* $P(X_2=1)=P(X_2=2)=P(X_2=3)=\frac{1}{3}$


*

*I know how to work with $X_1, X_2$ but I have no clue how to do anything with $Y, Z$. Why $Y\sim X_1$ and $Z\sim X_2$?

*How do I approach the question: Calculate $P(Y=y, Z=z)$?
 A: Assume that $X_1$ and $X_2$ are independent. The joint probability function of $X_1,X_2$ is given by
$$
\begin{array}{c|lcr}
X_2/X_1 & 0 & 1 & 2 \\
\hline
1 & 1/12 & 1/6 & 1/12 \\
2 & 1/12 & 1/6 & 1/12 \\
3 & 1/12 & 1/6 & 1/12
\end{array}
$$
The table for $Y=\min\{X_1,X_2\}$ is
$$
\begin{array}{c|lcr}
X_2/X_1 & 0 & 1 & 2 \\
\hline
1 & 0 & 1 & 1 \\
2 & 0 & 1 & 2 \\
3 & 0 & 1 & 2
\end{array}
$$
So $P(Y=0)=1/4$, $P(Y=1)=7/12$, $P(Y=2)=1/6$. The table for $Z=\max\{X_1,X_2\}$ is 
$$
\begin{array}{c|lcr}
X_2/X_1 & 0 & 1 & 2 \\
\hline
1 & 1 & 1 & 2 \\
2 & 2 & 2 & 2 \\
3 & 3 & 3 & 3
\end{array}
$$
So $P(Z=1)=1/4$, $P(Z=2)=5/12$ and $P(Z=3)=1/3$.
For the joint probability of $Y,Z$,$P(Y=y,Z=z)$, where $y\leq z$, analyze by cases: 
$$Y=0,Z=1\leftrightarrow X_1=0,X_2=1$$
$$Y=0,Z=2\leftrightarrow X_1=0,X_2=2$$
$$Y=0,Z=3\leftrightarrow X_1=0,X_2=3$$
$$Y=1,Z=1\leftrightarrow X_1=1,X_2=1$$
$$Y=1,Z=2\leftrightarrow X_1=1,X_2=2\;\text{or}\;X_1=2,X_2=1$$
$$\vdots$$
Using independence yo get the joint probability
$$
\begin{array}{c|lcr}
Z/Y & 0 & 1 & 2 \\
\hline
1 & 1/12 & 1/6 & 0 \\
2 & 1/12 & 1/4 & 1/12 \\
3 & 1/12 & 1/6 & 1/12
\end{array}
$$
A: You have not said what the joint distribution of $X_1,X_2$ is. But if we assume they are independent, although you didn't say that, then the joint distribution is as follows:
$$
\begin{array}{c|ccc|c}
_{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\
\hline 0 & 1/12 & 1/12 & 1/12 \\
1 & 1/6 & 1/6 & 1/6 \\
2 & 1/12 & 1/12 & 1/12 \\
\hline
\end{array}
$$
The values of $\max\{X_1,X_2\}$ are as follows:
$$
\begin{array}{c|ccc|c}
_{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\
\hline 0 & 0 & 1 & 2 \\
1 & 1 & 1 & 2 \\
2 & 2 & 2 & 2 \\
\hline
\end{array}
$$
Therefore
\begin{align}
\Pr(\max = 2) & = \frac 1 {12} + \frac 1 {12} + \frac 1 {12} + \frac 1 6 + \frac 1 {12} & & = \frac 1 2, \\[10pt]
\Pr(\max = 1) & = \frac 1 6 + \frac 1 6 + \frac 1 {12} & & = \frac 5 {12}, \\[10pt]
\Pr(\max = 0) & = \frac 1 {12} & & = \frac 1 {12}.
\end{align}
Similarly the values of $\min\{X_1,X_2\}$ are as follows:
$$
\begin{array}{c|ccc|c}
_{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\
\hline 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 1 \\
2 & 0 & 1 & 2 \\
\hline
\end{array}
$$
and you can find the probabilities in the same way.
