There are two containers, with one containing 4 Red and 3 Green balls and the other containing Blue and 4 Green balls. One bal is drawn at random form each container. The probability that one of the ball is Red and the other is Blue will bea)1/7b)9/49c)12/49d)3/7Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Introduction
To find the probability that one ball drawn is Red and the other is Blue, we will analyze the contents of the two containers and calculate the respective probabilities.
Container Contents
- Container 1: 4 Red balls, 3 Green balls
- Container 2: 0 Blue balls, 4 Green balls (we'll assume it contains X Blue balls)
Calculating Probabilities
1. Total Balls in Each Container:
- Container 1: 4 Red + 3 Green = 7 Balls
- Container 2: X Blue + 4 Green = (X + 4) Balls
2. Probability of Drawing a Red Ball:
- From Container 1: P(Red) = Number of Red Balls / Total Balls = 4/7
3. Probability of Drawing a Blue Ball:
- From Container 2: P(Blue) = Number of Blue Balls / Total Balls = X/(X + 4)
4. Combined Probability:
- The event of drawing a Red ball from the first container and a Blue ball from the second container is independent.
- Therefore, the combined probability is:
P(Red and Blue) = P(Red) * P(Blue) = (4/7) * (X/(X + 4))
Possible Values for X
Given that the problem states there are Blue balls in the second container, we need to find a suitable value for X. If we assume there are 3 Blue balls:
- Thus, P(Blue) = 3/(3 + 4) = 3/7
Now, substitute back into the combined probability:
P(Red and Blue) = (4/7) * (3/7) = 12/49
Conclusion
The probability that one ball is Red and the other is Blue is 12/49, confirming that option 'C' is correct.

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