Bag $$A$$ contains 3 white, 7 red balls and bag $$B$$ contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag $$A$$, if the ball drawn is white, is:

$$P(A) = P(B) = 1/2$$. $$P(W|A) = 3/10$$, $$P(W|B) = 3/5$$.

$$P(W) = P(A) \cdot P(W|A) + P(B) \cdot P(W|B) = \frac{1}{2} \cdot \frac{3}{10} + \frac{1}{2} \cdot \frac{3}{5} = \frac{3}{20} + \frac{3}{10} = \frac{3}{20} + \frac{6}{20} = \frac{9}{20}$$

By Bayes' theorem: $$P(A|W) = \frac{P(A) \cdot P(W|A)}{P(W)} = \frac{3/20}{9/20} = \frac{3}{9} = \frac{1}{3}$$

The answer is Option (3): $$\boxed{\frac{1}{3}}$$.