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A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession. The probability that one of them is black and the other is white is
- a)2/3
- b)4/5
- c)1/2
- d)1/3
Correct answer is option 'C'. Can you explain this answer?
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A bag contains 10 white balls and 15 black balls. Two balls are drawn ...
Bag contains 10 white balls and 15 black balls. Required probability
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A bag contains 10 white balls and 15 black balls. Two balls are drawn ...
Solution:
Given that there are 10 white balls and 15 black balls in the bag.
We need to find the probability that one of them is black and the other is white.
Let us consider the following cases:
Case 1: First ball is white and second ball is black
The probability of the first ball being white is 10/25 = 2/5
The probability of the second ball being black, given that the first ball was white is 15/24 = 5/8
Therefore, the probability of getting one white ball and one black ball in this case is (2/5) * (5/8) = 1/4
Case 2: First ball is black and second ball is white
The probability of the first ball being black is 15/25 = 3/5
The probability of the second ball being white, given that the first ball was black is 10/24 = 5/12
Therefore, the probability of getting one white ball and one black ball in this case is (3/5) * (5/12) = 1/4
Hence, the total probability of getting one white ball and one black ball is the sum of the probabilities in both cases:
P = 1/4 + 1/4 = 1/2
Therefore, the required probability is 1/2, which is option (c).
Given that there are 10 white balls and 15 black balls in the bag.
We need to find the probability that one of them is black and the other is white.
Let us consider the following cases:
Case 1: First ball is white and second ball is black
The probability of the first ball being white is 10/25 = 2/5
The probability of the second ball being black, given that the first ball was white is 15/24 = 5/8
Therefore, the probability of getting one white ball and one black ball in this case is (2/5) * (5/8) = 1/4
Case 2: First ball is black and second ball is white
The probability of the first ball being black is 15/25 = 3/5
The probability of the second ball being white, given that the first ball was black is 10/24 = 5/12
Therefore, the probability of getting one white ball and one black ball in this case is (3/5) * (5/12) = 1/4
Hence, the total probability of getting one white ball and one black ball is the sum of the probabilities in both cases:
P = 1/4 + 1/4 = 1/2
Therefore, the required probability is 1/2, which is option (c).
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