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An urn contains 2 red and 1 green balls,  another 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 first urn  is 
  • a)
    4/7
  • b)
    3/7
  • c)
    2/3
  • d)
    7/12
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
An urn contains 2 red and 1 green balls, another urn contains 2 red an...
Given:
- Urn 1 contains 2 red and 1 green balls.
- Urn 2 contains 2 red and 2 green balls.
- A ball is drawn from a randomly selected urn.
- The drawn ball is red.

To find:
The probability that the ball was drawn from the first urn.

Solution:
Let's define the following events:
- R1: a ball is drawn from the first urn.
- R2: a ball is drawn from the second urn.
- R: a red ball is drawn.
We need to find P(R1|R), i.e. the probability that the ball was drawn from the first urn given that a red ball was drawn.

Using Bayes' theorem, we have:
P(R1|R) = P(R|R1) * P(R1) / P(R)
where:
- P(R|R1) is the probability of drawing a red ball given that the first urn was selected. This is simply the proportion of red balls in the first urn, which is 2/3.
- P(R1) is the prior probability of selecting the first urn, which is 1/2 since either urn could have been selected.
- P(R) is the total probability of drawing a red ball, which can be calculated using the law of total probability:
P(R) = P(R|R1) * P(R1) + P(R|R2) * P(R2)
where:
- P(R|R2) is the probability of drawing a red ball given that the second urn was selected. This is the proportion of red balls in the second urn, which is 2/4 or 1/2.
- P(R2) is the prior probability of selecting the second urn, which is also 1/2.

Substituting the values, we get:
P(R1|R) = (2/3 * 1/2) / [(2/3 * 1/2) + (1/2 * 1/2)]
= 4/7

Therefore, the probability that the ball was drawn from the first urn given that a red ball was drawn is 4/7.
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Community Answer
An urn contains 2 red and 1 green balls, another urn contains 2 red an...
Total no.of balls=7
total no. of red balls =4
so,the probability of red balls =total no. of red balls /total no. of balls
=4/7
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An urn contains 2 red and 1 green balls, another 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 first urn isa)4/7b)3/7c)2/3d)7/12Correct answer is option 'A'. Can you explain this answer?
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An urn contains 2 red and 1 green balls, another 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 first urn isa)4/7b)3/7c)2/3d)7/12Correct answer is option 'A'. 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, another 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 first urn isa)4/7b)3/7c)2/3d)7/12Correct answer is option 'A'. 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, another 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 first urn isa)4/7b)3/7c)2/3d)7/12Correct answer is option 'A'. Can you explain this answer?.
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