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An urn contains 2 red and 1 green balls. nother urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______.
- a)4/7
- b)3/7
- c)2/3
- d)7/12
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An urn contains 2 red and 1 green balls. nother urn contains 2 red and...
Solution:
Given data:
- Urn 1: 2 red and 1 green balls
- Urn 2: 2 red and 2 green balls
- A ball is drawn randomly and found to be red
Let's find out the probability of drawing a red ball from each urn.
Probability of drawing a red ball from Urn 1:
= (number of red balls in Urn 1) / (total number of balls in Urn 1)
= 2 / 3
Probability of drawing a red ball from Urn 2:
= (number of red balls in Urn 2) / (total number of balls in Urn 2)
= 2 / 4
= 1 / 2
Now, let's apply Bayes' theorem to find the probability of drawing a ball from Urn 1 given that the ball drawn is red.
Bayes' Theorem:
P(Urn 1 | red) = P(red | Urn 1) * P(Urn 1) / P(red)
where,
P(Urn 1 | red) = probability of drawing from Urn 1 given that the ball drawn is red
P(red | Urn 1) = probability of drawing a red ball from Urn 1 (calculated above)
P(Urn 1) = probability of selecting Urn 1 (given as 1/2)
P(red) = probability of drawing a red ball (calculated below)
Probability of drawing a red ball:
P(red) = P(red | Urn 1) * P(Urn 1) + P(red | Urn 2) * P(Urn 2)
= (2/3) * (1/2) + (1/2) * (1/2)
= 7/12
Substituting the values in Bayes' theorem formula:
P(Urn 1 | red) = (2/3) * (1/2) / (7/12)
= 4/7
Therefore, the probability that the ball was drawn from Urn 1 given that it is red is 4/7.
Hence, option (a) 4/7 is the correct answer.
Given data:
- Urn 1: 2 red and 1 green balls
- Urn 2: 2 red and 2 green balls
- A ball is drawn randomly and found to be red
Let's find out the probability of drawing a red ball from each urn.
Probability of drawing a red ball from Urn 1:
= (number of red balls in Urn 1) / (total number of balls in Urn 1)
= 2 / 3
Probability of drawing a red ball from Urn 2:
= (number of red balls in Urn 2) / (total number of balls in Urn 2)
= 2 / 4
= 1 / 2
Now, let's apply Bayes' theorem to find the probability of drawing a ball from Urn 1 given that the ball drawn is red.
Bayes' Theorem:
P(Urn 1 | red) = P(red | Urn 1) * P(Urn 1) / P(red)
where,
P(Urn 1 | red) = probability of drawing from Urn 1 given that the ball drawn is red
P(red | Urn 1) = probability of drawing a red ball from Urn 1 (calculated above)
P(Urn 1) = probability of selecting Urn 1 (given as 1/2)
P(red) = probability of drawing a red ball (calculated below)
Probability of drawing a red ball:
P(red) = P(red | Urn 1) * P(Urn 1) + P(red | Urn 2) * P(Urn 2)
= (2/3) * (1/2) + (1/2) * (1/2)
= 7/12
Substituting the values in Bayes' theorem formula:
P(Urn 1 | red) = (2/3) * (1/2) / (7/12)
= 4/7
Therefore, the probability that the ball was drawn from Urn 1 given that it is red is 4/7.
Hence, option (a) 4/7 is the correct answer.
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Community Answer
An urn contains 2 red and 1 green balls. nother urn contains 2 red and...
Probability of getting red urn from urn one=no.of red urn in urn first divideby total no. of urn in urn one
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An urn contains 2 red and 1 green balls. nother urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______.a)4/7b)3/7c)2/3d)7/12Correct answer is option 'C'. Can you explain this answer?
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An urn contains 2 red and 1 green balls. nother urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______.a)4/7b)3/7c)2/3d)7/12Correct answer is option 'C'. Can you explain this answer? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about An urn contains 2 red and 1 green balls. nother urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______.a)4/7b)3/7c)2/3d)7/12Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An urn contains 2 red and 1 green balls. nother urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______.a)4/7b)3/7c)2/3d)7/12Correct answer is option 'C'. Can you explain this answer?.
An urn contains 2 red and 1 green balls. nother urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______.a)4/7b)3/7c)2/3d)7/12Correct answer is option 'C'. Can you explain this answer? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about An urn contains 2 red and 1 green balls. nother urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______.a)4/7b)3/7c)2/3d)7/12Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An urn contains 2 red and 1 green balls. nother urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______.a)4/7b)3/7c)2/3d)7/12Correct answer is option 'C'. Can you explain this answer?.
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