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There are two urns. The first urn contains 3 red and 5 white balls whereas the second urn contains 4 red and 6 white balls. A ball is taken at random from the first urn and is transferred to the second urn. Now another ball is selected at random from the second arm. The probability that the second ball would be red is
- a)7/20
- b)35/88
- c)17/52
- d)3/20
Correct answer is option 'B'. Can you explain this answer?
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There are two urns. The first urn contains 3 red and 5 white balls whe...
Solution:
Given,
First urn contains 3 red and 5 white balls
Second urn contains 4 red and 6 white balls
A ball is taken at random from the first urn and is transferred to the second urn.
Probability of selecting a red ball from the first urn is: P(R1) = 3/8
Probability of selecting a white ball from the first urn is: P(W1) = 5/8
After transferring a ball from first urn to second urn, the number of balls in the second urn is 11.
Probability of selecting a red ball from the second urn, given that a red ball was transferred from the first urn is:
P(R2/R1) = (4+1)/(11+1) = 5/12
Probability of selecting a white ball from the second urn, given that a red ball was transferred from the first urn is:
P(W2/R1) = (6)/(11+1) = 1/6
Probability of selecting a red ball from the second urn can be obtained using the law of total probability as follows:
P(R2) = P(R2/R1)P(R1) + P(R2/W1)P(W1)
P(R2) = (5/12)(3/8) + (4/11)(1/6)
P(R2) = 35/88
Therefore, the probability that the second ball would be red is 35/88.
Given,
First urn contains 3 red and 5 white balls
Second urn contains 4 red and 6 white balls
A ball is taken at random from the first urn and is transferred to the second urn.
Probability of selecting a red ball from the first urn is: P(R1) = 3/8
Probability of selecting a white ball from the first urn is: P(W1) = 5/8
After transferring a ball from first urn to second urn, the number of balls in the second urn is 11.
Probability of selecting a red ball from the second urn, given that a red ball was transferred from the first urn is:
P(R2/R1) = (4+1)/(11+1) = 5/12
Probability of selecting a white ball from the second urn, given that a red ball was transferred from the first urn is:
P(W2/R1) = (6)/(11+1) = 1/6
Probability of selecting a red ball from the second urn can be obtained using the law of total probability as follows:
P(R2) = P(R2/R1)P(R1) + P(R2/W1)P(W1)
P(R2) = (5/12)(3/8) + (4/11)(1/6)
P(R2) = 35/88
Therefore, the probability that the second ball would be red is 35/88.
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There are two urns. The first urn contains 3 red and 5 white balls whereas the second urn contains 4 red and 6 white balls. A ball is taken at random from the first urn and is transferred to the second urn. Now another ball is selected at random from the second arm. The probability that the second ball would be red isa)7/20b)35/88c)17/52d)3/20Correct answer is option 'B'. Can you explain this answer?
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Solutions for There are two urns. The first urn contains 3 red and 5 white balls whereas the second urn contains 4 red and 6 white balls. A ball is taken at random from the first urn and is transferred to the second urn. Now another ball is selected at random from the second arm. The probability that the second ball would be red isa)7/20b)35/88c)17/52d)3/20Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for CA Foundation.
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