The correct answer is $\boxed{\frac{12}{49}}$.
The probability of event A happening, given that event B has already happened, is called the conditional probability of A given B, and is denoted by $P(A|B)$. It can be calculated using the following formula:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
In this case, event A is “the first ball drawn is red” and event B is “the second ball drawn is blue”. We are asked to find the probability of event A happening, given that event B has already happened.
The probability of event A happening is the number of ways to draw a red ball from the first container and a blue ball from the second container, divided by the total number of ways to draw two balls from the two containers. This is:
$$P(A) = \frac{4 \times 3}{{7 \times 8}} = \frac{3}{28}$$
The probability of event B happening is the number of ways to draw a blue ball from the second container, divided by the total number of ways to draw a ball from the second container. This is:
$$P(B) = \frac{3}{{7}}$$
Therefore, the probability of event A happening, given that event B has already happened, is:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{4 \times 3}{{7 \times 8}} \times \frac{3}{{7}}}{\frac{3}{{7}}} = \frac{12}{49}$$