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. A ball is drawn at random from the bag, the number on the ball is noted, and the ball is returned. This procedure is repeated 10 times, and let $Y$ be the sum of the 10 numbers observed. The following is the process of finding the mean $\mathrm { E } ( Y )$ and variance $\mathrm { V } ( Y )$ of the random variable $Y$.

Consider the 60 balls in the bag as a population. When a ball is drawn at random from this population, let $X$ be the random variable representing the number on the ball. The probability distribution of $X$, which 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 $\bar { X }$,

$$\mathrm { E } ( \bar { X } ) = \frac { 7 } { 3 } , \quad \mathrm {~V} ( \bar { X } ) = \text { (나) }$$

If the number on the $n$-th ball drawn from the bag is $X _ { n }$, then

$$Y = \sum _ { n = 1 } ^ { 10 } X _ { n } = 10 \bar { X }$$

so

$$\mathrm { E } ( Y ) = \frac { 70 } { 3 } , \quad \mathrm {~V} ( Y ) = \text { (다) }$$

If 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) $\frac { 13 } { 2 }$