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There are two containers, with one containing 4 Red and 3 Green balls and the other containing 3 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 be       

  • a)
    1/7

  • b)
    9/49

  • c)
    12/49

  • d)
    3/7

Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
There are two containers, with one containing 4 Red and 3 Green balls ...
Problem Analysis

To solve this problem, we need to find the probability of drawing one Red ball and one Blue ball from two different containers. Let's analyze the given information:

- Container 1: 4 Red balls and 3 Green balls
- Container 2: 3 Blue balls and 4 Green balls

Solution

Step 1: Find the probability of drawing a Red ball from Container 1

The probability of drawing a Red ball from Container 1 can be calculated as the ratio of the number of Red balls to the total number of balls in Container 1.
Probability of drawing a Red ball from Container 1 = 4/7

Step 2: Find the probability of drawing a Blue ball from Container 2

Similarly, the probability of drawing a Blue ball from Container 2 can be calculated as the ratio of the number of Blue balls to the total number of balls in Container 2.
Probability of drawing a Blue ball from Container 2 = 3/7

Step 3: Find the probability of drawing one Red and one Blue ball

Now, we need to find the probability of drawing one Red ball from Container 1 and one Blue ball from Container 2. Since the two events are independent, we can multiply the probabilities of each event.
Probability of drawing one Red and one Blue ball = Probability of drawing a Red ball from Container 1 × Probability of drawing a Blue ball from Container 2

Probability of drawing one Red and one Blue ball = (4/7) × (3/7) = 12/49

Therefore, the probability that one of the balls is Red and the other is Blue is 12/49.

Conclusion

The correct answer is option 'C', 12/49.
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Community Answer
There are two containers, with one containing 4 Red and 3 Green balls ...
P_red=4÷(4+3)=4/7,P_blue=3÷(3+4)=3/7,So,Resultant probability=(4/7)×(3/7)=12/49.
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There are two containers, with one containing 4 Red and 3 Green balls and the other containing 3 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 be a)1/7b)9/49c)12/49d)3/7Correct answer is option 'C'. Can you explain this answer?
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