A bag contains 10 balls labeled with the number 1, 20 balls labeled with the number 2, and 30 balls labeled with the number 3. One ball is randomly drawn from the bag, the number on the ball is confirmed, and then the ball is returned. This procedure is repeated 10 times, and the sum of the 10 numbers confirmed is the random variable $Y$. The following is the process of finding the mean $\mathrm { E } ( Y )$ and variance $\mathrm { V } ( Y )$ of the random variable $Y$.

Let the 60 balls in the bag be the population. When one ball is randomly drawn from this population, let the number on the ball be the random variable $X$. The probability distribution of $X$, that is, the probability distribution of the population, is shown in the following table.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$X$ & 1 & 2 & 3 & Total \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 3 }$ & $\frac { 1 } { 2 }$ & 1 \\
\hline
\end{tabular}
\end{center}

Therefore, the population mean $m$ and population variance $\sigma ^ { 2 }$ are
$$m = \mathrm { E } ( X ) = \frac { 7 } { 3 } , \quad \sigma ^ { 2 } = \mathrm { V } ( X ) = \text { (가) }$$

When a sample of size 10 is randomly extracted from the population and the sample mean is $\bar { X }$,
$$\mathrm { E } ( \bar { X } ) = \frac { 7 } { 3 } , \quad \mathrm {~V} ( \bar { X } ) = \text { (나) }$$

Let the number on the $n$-th ball drawn from the bag be $X _ { n }$. Then
$$Y = \sum _ { n = 1 } ^ { 10 } X _ { n } = 10 \bar { X }$$
so
$$\mathrm { E } ( Y ) = \frac { 70 } { 3 } , \quad \mathrm {~V} ( Y ) = \text { (다) }$$

When the numbers that fit (가), (나), and (다) are $p , q , r$ respectively, what is the value of $p + q + r$? [4 points]\\
(1) $\frac { 31 } { 6 }$\\
(2) $\frac { 11 } { 2 }$\\
(3) $\frac { 35 } { 6 }$\\
(4) $\frac { 37 } { 6 }$\\
(5) $\frac { 13 } { 2 }$