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Box A contains 2 black and 3 red balls, while box B contains 3 black and 4 red balls. Out of these two boxes one is selected at random, and the probability of choosing box A is double that of box B. If a red ball is drawn from the selected box then the probability that it has come from box B is
- a)21/41
- b)10/31
- c)12/31
- d)13/41
Correct answer is option 'B'. Can you explain this answer?
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Box A contains 2 black and 3 red balls, while box B contains 3 black a...
Solution:
Given: Box A contains 2 black and 3 red balls, while box B contains 3 black and 4 red balls.
Let P(A) and P(B) be the probabilities of selecting box A and box B, respectively.
Given: The probability of choosing box A is double that of box B.
So, P(A) = 2P(B)
Let E be the event of drawing a red ball from the selected box.
We need to find the probability that the red ball has come from box B given that it is drawn.
Let P(E|A) and P(E|B) be the conditional probabilities of drawing a red ball from box A and box B, respectively.
P(E|A) = 3/5, as box A contains 3 red balls out of 5 balls in total.
P(E|B) = 4/7, as box B contains 4 red balls out of 7 balls in total.
Using Bayes' theorem, we can find the required probability:
P(B|E) = P(E|B) * P(B) / [P(E|A) * P(A) + P(E|B) * P(B)]
P(A) = 2P(B)
=> P(B) = P(A) / 2
=> P(B) = 2/3 and P(A) = 4/3
Putting the values in Bayes' theorem:
P(B|E) = (4/7) * (2/3) / [(3/5) * (4/3) + (4/7) * (2/3)]
P(B|E) = 10/31
Therefore, the probability that the red ball has come from box B given that it is drawn is 10/31.
Hence, option (B) is the correct answer.
Given: Box A contains 2 black and 3 red balls, while box B contains 3 black and 4 red balls.
Let P(A) and P(B) be the probabilities of selecting box A and box B, respectively.
Given: The probability of choosing box A is double that of box B.
So, P(A) = 2P(B)
Let E be the event of drawing a red ball from the selected box.
We need to find the probability that the red ball has come from box B given that it is drawn.
Let P(E|A) and P(E|B) be the conditional probabilities of drawing a red ball from box A and box B, respectively.
P(E|A) = 3/5, as box A contains 3 red balls out of 5 balls in total.
P(E|B) = 4/7, as box B contains 4 red balls out of 7 balls in total.
Using Bayes' theorem, we can find the required probability:
P(B|E) = P(E|B) * P(B) / [P(E|A) * P(A) + P(E|B) * P(B)]
P(A) = 2P(B)
=> P(B) = P(A) / 2
=> P(B) = 2/3 and P(A) = 4/3
Putting the values in Bayes' theorem:
P(B|E) = (4/7) * (2/3) / [(3/5) * (4/3) + (4/7) * (2/3)]
P(B|E) = 10/31
Therefore, the probability that the red ball has come from box B given that it is drawn is 10/31.
Hence, option (B) is the correct answer.
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Box A contains 2 black and 3 red balls, while box B contains 3 black and 4 red balls. Out of these two boxes one is selected at random, and the probability of choosing box A is double that of box B. If a red ball is drawn from the selected box then the probability that it has come from box B isa)21/41b)10/31c)12/31d)13/41Correct answer is option 'B'. Can you explain this answer?
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