Let the number of red marbles be $\(r\)$.
Then the number of blue marbles is $\(2 r\)$ (twice red).
Let the number of green marbles be $\(g\)$, where $\(g \geq 1\)$.
The total number of marbles is:

$$
\(r+2 r+g=3 r+g\)
$$


The fraction of red marbles is:

$$
\(\frac{r}{3 r+g}\)
$$


Since $\(g \geq 1\)$, the denominator is at least $\(3 r+1\)$.
Hence:

$$
\(\frac{r}{3 r+g} \leq \frac{r}{3 r+1}\)
$$


For large $r$, this ratio approaches $\(\frac{r}{3 r}=\frac{1}{3}\)$ from below.
If $\(r=1\)$, then:

$$
\(\frac{1}{3(1)+g}=\frac{1}{3+g}\)
$$

which is less than $\(\frac{1}{3}\)$ for any positive $g$.
Therefore, the ratio $\(\frac{\text { red marbles }}{\text { total marbles }}\)$ is always less than $\(\frac{1}{3}\)$.

Answer:
A) Less than $\(\frac{1}{3}\)$