Answer
Sample Space: $\color{blue}{\begin{array}{cc}
\underline{(D_1,D_2)} & \rm Die\ 2\ (D_2) \\
\rm Die\ 1\ (D_1)& \begin{array}{|c|c|c|c|c|c|} \hline
D_1\backslash D_2& 1 & 2 & 2 & 3 & 3 & 4\\ \hline
1 & (1,1) & (1,2) & (1,2) & (1,3) & (1,3) & (1,4) \\
3 & (3,1) & (3,2) & (3,2) & (3,3) & (3,3) & (3,4) \\
4 & (4,1) & (4,2) & (4,2) & (4,3) & (4,3) & (4,4) \\
5 & (5,1) & (5,2) & (5,2) & (5,3) & (5,3) & (5,4) \\
6 & (6,1) & (6,2) & (6,2) & (6,3) & (6,3) & (6,4) \\
8 & (8,1) & (8,2) & (8,2) & (8,3) & (8,3) & (8,4) \\ \hline
\end{array}
\end{array}}$
pdf of $X$:
$\color{blue}{p_X(k) = \begin{cases} 1/36, & k=2,12 \\ 2/36, & k=3,11 \\ 3/36, & k=4,10 \\ 4/36, & k=5,9 \\ 5/36, & k=6,8 \\ 6/36, & k=7 \end{cases}}$
which is the same as that of the total number of spots in the upturned faces in a roll of two ordinary dice.
Work Step by Step
$\underline{\rm The\ Sample\ Space}$
Since each of Die 1 and Die 2 has a total of six possible outcomes, then the total number of possible outcomes, when both dice are rolled, is $6\cdot 6 = 36$. These 36 elements of the sample space are shown in the body of the table below.
$\begin{array}{cc}
\underline{(D_1,D_2)} & \rm Die\ 2\ (D_2) \\
\rm Die\ 1\ (D_1)& \begin{array}{|c|c|c|c|c|c|} \hline
D_1\backslash D_2& 1 & 2 & 2 & 3 & 3 & 4\\ \hline
1 & (1,1) & (1,2) & (1,2) & (1,3) & (1,3) & (1,4) \\
3 & (3,1) & (3,2) & (3,2) & (3,3) & (3,3) & (3,4) \\
4 & (4,1) & (4,2) & (4,2) & (4,3) & (4,3) & (4,4) \\
5 & (5,1) & (5,2) & (5,2) & (5,3) & (5,3) & (5,4) \\
6 & (6,1) & (6,2) & (6,2) & (6,3) & (6,3) & (6,4) \\
8 & (8,1) & (8,2) & (8,2) & (8,3) & (8,3) & (8,4) \\ \hline
\end{array}
\end{array}$
$\underline{{\rm The\ Random\ Variable}\ X}$
The sums of the elements of each ordered pair (the total number of spots showing on the upturned faces of the two dice) shown in the table above are shown in the corresponding cell in the body of the table below.
$\begin{array}{cc}
\underline{D_1+D_2} & {\rm Die\ 2}\ (D_2) \\
{\rm Die\ 1}\ (D_1) & \begin{array}{|c|c|c|c|c|c|} \hline
D_1\backslash D_2& 1 & 2 & 2 & 3 & 3 & 4\\ \hline
1 & 2 & 3 & 3 & 4 & 4 & 5 \\
3 & 4 & 5 & 5 & 6 & 6 & 7 \\
4 & 5 & 6 & 6 & 7 & 7 & 8 \\
5 & 6 & 7 & 7 & 8 & 8 & 9 \\
6 & 7 & 8 & 8 & 9 & 9 & 10 \\
8 & 9 & 10 & 10 & 11 & 11 & 12 \\ \hline
\end{array}
\end{array}$
The frequency of the different sums (total number of spots in the upturned faces) that occur among the 36 sums that appear in the table directly above are
$\begin{array}{|c|c|} \hline
\text{Sum} & \text{Frequency} \\ \hline
2 & 1 \\
3 & 2 \\
4 & 3 \\
5 & 4 \\
6 & 5 \\
7 & 6 \\
8 & 5 \\
9 & 4 \\
10 & 3 \\
11 & 2 \\
12 & 1 \\ \hline
\end{array}$
Thus, if we let the random variable (r.v.) $X=D_1+D_2$, that is $X$ is the r.v. that denotes the total number of spots in the upturned faces in a roll of the two dice, then the pdf of $X$ is
$\color{blue}{p_X(k) = \begin{cases} 1/36, & k=2,12 \\ 2/36, & k=3,11 \\ 3/36, & k=4,10 \\ 4/36, & k=5,9 \\ 5/36, & k=6,8 \\ 6/36, & k=7 \end{cases}}$
which is precisely the pdf for the random variable denoting the total number of spots on the upturned faces in a roll of two ordinary dice as shown in Table 3.3.3 (Example 3.3.6, p. 123).
