This question asks us to find the probability of a specific combined event: drawing one red ball from Bag I AND one blue ball from Bag II. The draws from each bag are independent events.

Analyzing Bag Contents

• Bag I: Contains 4 red balls and 3 green balls. The total number of balls in Bag I is $4 + 3 = 7$.
• Bag II: Contains 3 blue balls and 4 green balls. The total number of balls in Bag II is $3 + 4 = 7$.

Calculating Individual Probabilities

We need to find the probability of the desired outcome for each bag separately.

• Probability of drawing a red ball from Bag I:
  • Number of red balls in Bag I = 4
  • Total balls in Bag I = 7
  • The probability, let's call it $P(\text{Red from Bag I})$, is $\frac{\text{Number of red balls}}{\text{Total balls}} = \frac{4}{7}$.
• Probability of drawing a blue ball from Bag II:
  • Number of blue balls in Bag II = 3
  • Total balls in Bag II = 7
  • The probability, let's call it $P(\text{Blue from Bag II})$, is $\frac{\text{Number of blue balls}}{\text{Total balls}} = \frac{3}{7}$.

Calculating Combined Probability for Independent Events

Since the event of drawing a ball from Bag I is independent of the event of drawing a ball from Bag II, the probability of both events occurring is the product of their individual probabilities.

The probability that one drawn ball is red (from Bag I) and the other is blue (from Bag II) is:

$ P(\text{Red from Bag I} \text{ and } \text{Blue from Bag II}) = P(\text{Red from Bag I}) \times P(\text{Blue from Bag II}) $

Substituting the calculated probabilities:

$ = \frac{4}{7} \times \frac{3}{7} $

$ = \frac{4 \times 3}{7 \times 7} $

$ = \frac{12}{49} $

Final Probability

Therefore, the probability that one of the drawn balls is red and the other is blue is $\frac{12}{49}$.