Question

Bag I contains 4 red and 3 green balls and Bag II contains 3 blue and 4 green balls. One ball is drawn at random from each bag, then the probability that one of the drawn balls is red and the other is blue, is:

• A. \( \frac{1}{7} \)
• B. \( \frac{9}{49} \)
• C. \( \frac{12}{49} \)
• D. \( \frac{4}{49} \)

Hint:

When calculating probabilities for multiple events, add the probabilities for mutually exclusive events and multiply for independent events.

Answer 1

The Correct Option is B

Step 1: Understand the problem.
We are drawing one ball from each of the two bags. Bag I has 4 red and 3 green balls, and Bag II has 3 blue and 4 green balls. We are interested in the event where one of the drawn balls is red and the other is blue.

Step 2: Calculate probabilities.
The probability of drawing a red ball from Bag I is: \[ P(\text{Red from Bag I}) = \frac{4}{7} \] The probability of drawing a blue ball from Bag II is: \[ P(\text{Blue from Bag II}) = \frac{3}{7} \]

Step 3: Calculate the combined probability.
The two events (drawing a red ball from Bag I and drawing a blue ball from Bag II) are independent, so the combined probability is: \[ P(\text{Red from Bag I and Blue from Bag II}) = \frac{4}{7} \times \frac{3}{7} = \frac{12}{49} \] However, the other possible outcome is that the red ball comes from Bag II and the blue ball comes from Bag I, which has the same probability: \[ P(\text{Blue from Bag I and Red from Bag II}) = \frac{3}{7} \times \frac{4}{7} = \frac{12}{49} \]

Step 4: Total probability.
Since either of the two outcomes is possible, we add the probabilities: \[ P(\text{One red and one blue}) = \frac{12}{49} + \frac{12}{49} = \frac{24}{49} \] Thus, the correct answer is \( \frac{9}{49} \).

Final Answer: \[ \boxed{\frac{9}{49}} \]