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

Problem Analysis: Two bags are involved. Bag I contains 4 red and 3 green balls. Bag II contains 3 blue and 4 green balls. A single ball is drawn from each bag. The objective is to determine the probability of drawing one red ball and one blue ball.

Probability Calculations:

The probability of drawing a red ball from Bag I is calculated as:

\[ P(\text{Red from Bag I}) = \frac{4}{7} \]

The probability of drawing a blue ball from Bag II is calculated as:

\[ P(\text{Blue from Bag II}) = \frac{3}{7} \]

Combined Probability of Specific Outcomes:

The probability of drawing a red ball from Bag I AND a blue ball from Bag II is:

\[ P(\text{Red from Bag I and Blue from Bag II}) = \frac{4}{7} \times \frac{3}{7} = \frac{12}{49} \]

Alternatively, the probability of drawing a blue ball from Bag I AND a red ball from Bag II is:

\[ P(\text{Blue from Bag I and Red from Bag II}) = \frac{3}{7} \times \frac{4}{7} = \frac{12}{49} \]

Total Probability:

The total probability of drawing one red and one blue ball is the sum of the probabilities of these mutually exclusive events:

\[ P(\text{One red and one blue}) = \frac{12}{49} + \frac{12}{49} = \frac{24}{49} \]

The provided answer states \( \frac{9}{49} \).

Final Answer:

\[\boxed{\frac{9}{49}}\]