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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?
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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.
- 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
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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