To find the probability that one ball is red and the other is black when one ball is drawn from each bag, we need to consider two scenarios:

Drawing a red ball from the first bag and a black ball from the second bag.
Drawing a black ball from the first bag and a red ball from the second bag.

Let's calculate the probability for each scenario:

Scenario 1: Drawing a red ball from the first bag and a black ball from the second bag.

The probability of drawing a red ball from the first bag (which has 3 red and 5 black balls) is $$\frac{3}{3+5} = \frac{3}{8}$$. The probability of drawing a black ball from the second bag (which has 6 red and 4 black balls) is $$\frac{4}{6+4} = \frac{4}{10} = \frac{2}{5}$$. Thus, the probability of this scenario is the product of these two probabilities, given by:

$$P(\text{Red from first and Black from second}) = \frac{3}{8} \times \frac{2}{5} = \frac{3 \times 2}{8 \times 5} = \frac{6}{40} = \frac{3}{20}$$

Scenario 2: Drawing a black ball from the first bag and a red ball from the second bag.

The probability of drawing a black ball from the first bag is $$\frac{5}{3+5} = \frac{5}{8}$$. The probability of drawing a red ball from the second bag is $$\frac{6}{6+4} = \frac{6}{10} = \frac{3}{5}$$. Thus, the probability of this scenario is:

$$P(\text{Black from first and Red from second}) = \frac{5}{8} \times \frac{3}{5} = \frac{5 \times 3}{8 \times 5} = \frac{3}{8}$$

To find the total probability that one ball is red and the other is black, we add the probabilities of these two scenarios:

$$P(\text{One Red and One Black}) = P(\text{Red from first and Black from second}) + P(\text{Black from first and Red from second})$$ $$= \frac{3}{20} + \frac{3}{8}$$

To add these fractions, we need a common denominator, which is 40 in this case. So, we convert $$\frac{3}{8}$$ to an equivalent fraction with a denominator of 40.

$$\frac{3}{8} = \frac{3 \times 5}{8 \times 5} = \frac{15}{40}$$

Now, we can add the fractions:

$$P(\text{One Red and One Black}) = \frac{3}{20} + \frac{15}{40}$$

Since $$\frac{3}{20}$$ is equivalent to $$\frac{6}{40}$$, we now have:

$$P(\text{One Red and One Black}) = \frac{6}{40} + \frac{15}{40} = \frac{21}{40}$$

Therefore, the probability that one ball is red and the other is black is $$\frac{21}{40}$$, which corresponds to Option B.

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