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Let $E_1$ be the event of choosing Bag I and $E_2$ be the event of choosing Bag II. Let $R$ be the event of drawing a red ball.
A die is thrown. Bag I is chosen if the number is less than 3 (i.e., 1 or 2). Thus, $P(E_1) = \frac{2}{6} = \frac{1}{3}$. Bag II is chosen otherwise (i.e., 3, 4, 5, or 6). Thus, $P(E_2) = \frac{4}{6} = \frac{2}{3}$.
Bag I has 3 red and 4 white balls (Total 7). So, $P(R|E_1) = \frac{3}{7}$. Bag II has 8 red and 6 white balls (Total 14). So, $P(R|E_2) = \frac{8}{14} = \frac{4}{7}$.
The total probability of drawing a red ball is given by: $$P(R) = P(E_1) \cdot P(R|E_1) + P(E_2) \cdot P(R|E_2)$$ $$P(R) = \left(\frac{1}{3} \cdot \frac{3}{7}\right) + \left(\frac{2}{3} \cdot \frac{4}{7}\right)$$ $$P(R) = \frac{3}{21} + \frac{8}{21} = \frac{11}{21}$$
Final Answer: 11/21