Answer
$\color{blue}{p_X(k) = \begin{cases} 1/25, & k=1 \\ 3/25, & k=2 \\ 5/25, & k=3 \\ 7/25, & k=4 \\ 9/25, & k=5 \end{cases}}$
$\color{blue}{p_V(k) =\begin{cases} 1/25, & k=2, 10 \\ 2/25, & k=3, 9 \\ 3/25, & k=4, 8 \\ 4/25, & k=5, 7 \\ 5/25, & k=6 \end{cases}} $
Work Step by Step
The total number of possible outcomes when two balls are drawn with replacement from an urn containing five (5) balls numbered $1, 2, 3, 4,$ and $5$, is $25$ as there are 5 possible outcomes for each of the first and second balls. The outcomes are summarized in the table below. The outcomes are equally likely and each has a probability of $1/25$ of occurring.
$\begin{array}{cc}
\underline{(B_1,B_2)} & \rm Ball\ 2\ (B_2) \\
\rm Ball\ 1\ (B_1)& \begin{array}{|c|c|c|c|c|} \hline
B_1\backslash B_2& 1 & 2 & 3 & 4 & 5 \\ \hline
1 & (1,1) & (1,2) & (1,3) & (1,4) & (1,5) \\
2 & (2,1) & (2,2) & (2,3) & (2,4) & (2,5) \\
3 & (3,1) & (3,2) & (3,3) & (3,4) & (3,5) \\
4 & (4,1) & (4,2) & (4,3) & (4,4) & (4,5) \\
5 & (5,1) & (5,2) & (5,3) & (5,4) & (5,5) \\ \hline
\end{array}
\end{array}$
(a)
The table below lists the maximum number appearing when two balls are drawn with replacement from the urn previously described. For example, if the outcome is $(B_1, B_2)=(2,3)$, that is, Ball 1 is 2 and Ball 2 is 3, then the maximum number appearing is $3$, and this is indicated in row 2, column 3, in the body of the table below. Note that the possible maxima are 1, 2, 3, 4, and 5.
$\begin{array}{cc}
\underline{\max\{B_1,B_2\}} & \rm Ball\ 2\ (B_2) \\
\rm Ball\ 1\ (B_1) & \begin{array}{|c|c|c|c|c|} \hline
B_1\backslash B_2 & 1 & 2 & 3 & 4 & 5 \\ \hline
1 & 1 & 2 & 3 & 4 & 5 \\
2 & 2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 3 & 4 & 5 \\
4 & 4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 & 5 \\ \hline
\end{array}
\end{array}$
Since $X=\max\{B_1, B_2\}$, it can be easily seen from the table above that of the $25$ possible equally-likely outcomes, $X=1$ for 1 outcome, $X=2$ for 3 outcomes, $X=3$ for 5 outcomes, $X=4$ for 7 outcomes, and $X=5$ for 9 outcomes. Thus, the pdf of $X$ is
$\begin{align*}
p_X(k) &= P(X=k),\ k=1,2,3,4,5 \\
&= P(\max\{B_1,B_2\} = k) \\
\color{blue}{p_X(k)} &\color{blue}{= \begin{cases} 1/25, & k=1 \\ 3/25, & k=2 \\ 5/25, & k=3 \\ 7/25, & k=4 \\ 9/25, & k=5 \end{cases}}
\end{align*}$
(b)
The table below lists all the possible sums of the numbers appearing on two balls that are drawn with replacement from the urn previously described. For example, if the outcome is $(B_1, B_2)=(2, 4)$, then the sum is $2+4=6$ (see row 2, column 4 in the body of the table below). Observe that the possible sums are 2, 3, 4, 5, 6, 7, 8, 9, and 10.
$\begin{array}{cc}
\underline{B_1+B_2} & \rm Ball\ 2\ (B_2) \\
\rm Ball\ 1\ (B_1) & \begin{array}{|c|c|c|c|c|} \hline
B_1\backslash B_2 & 1 & 2 & 3 & 4 & 5 \\ \hline
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 3 & 4 & 5 & 6 & 7 \\
3 & 4 & 5 & 6 & 7 & 8 \\
4 & 5 & 6 & 7 & 8 & 9 \\
5 & 6 & 7 & 8 & 9 & 10 \\ \hline
\end{array}
\end{array}$
With $V = B_1+B_2$, it can be readily seen from the table above that of the $25$ possible equally-likely outcomes, $V=2$ for 1 outcome, $V=3$ for 2 outcomes, $V=4$ for 3 outcomes, $V=5$ for 4 outcomes, $V=6$ for 5 outcomes, $V=7$ for 4 outcomes, $V=8$ for 3 outcomes, $V=9$ for 2 outcomes, and $V=10$ for 1 outcome. Thus, the pdf of $V$ is
$\begin{align*}
p_V(k) &= P(V=k),\ k= 2,3,4,5,6,7,8,9,10 \\
&= P(B_1+B_2=k) \\
& \\
&= \begin{cases} 1/25, & k=2 \\ 2/25, & k=3 \\ 3/25, & k=4 \\ 4/25, & k=5 \\ 5/25, & k=6 \\ 4/25, & k=7 \\ 3/25, & k=8 \\ 2/25, & k=9 \\ 1/25, & k=10 \end{cases} \\
& \\
\color{blue}{p_V(k)} &=\color{blue}{\begin{cases} 1/25, & k=2, 10 \\ 2/25, & k=3, 9 \\ 3/25, & k=4, 8 \\ 4/25, & k=5, 7 \\ 5/25, & k=6 \end{cases}}
\end{align*}$
