Total Probability
An urn holds four red and six blue balls, and one ball drawn at the first stage is set aside without being seen before a second ball is drawn; what is the probability the second ball is red?
Select the correct option:
Solution
2/5
The law of total probability decomposes an event over a partition of mutually exclusive cases, here the unseen first ball being red or blue, via P(B) = ΣP(B | Aᵢ)P(Aᵢ). This conditioning-on-the-first-draw technique is a recurring JEE Advanced device. The first ball is red with probability 4/10 and blue with probability 6/10. If the first ball was red, the urn has 3 red of 9 remaining, so the second is red with probability 3/9; if the first was blue, the urn has 4 red of 9, giving 4/9. Combining, P(second red) = (4/10)(3/9) + (6/10)(4/9) = (12 + 24)/90 = 36/90 = 2/5. Option 1/3 mistakenly conditions on only the red-first branch. Option 4/9 uses only the blue-first branch. Option 3/10 forgets to renormalise over the reduced count of nine. The result equals the original red fraction 4/10 = 2/5, illustrating the symmetry that without information, position does not change marginal probability. This is a deep and often surprising point: because the discarded first ball is never observed, the second draw is statistically indistinguishable from simply drawing a single ball from the original urn, so its red probability must coincide with the initial proportion. Any other answer would imply that merely relabelling which draw we call second could alter chances, which symmetry forbids. Plausibility check: 2/5 matches the proportion of red balls initially, exactly as expected when the discarded ball is unobserved, and it lies sensibly between the two branch probabilities 3/9 and 4/9, confirming the total-probability computation.
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About This Question
- Subject
- mathematics
- Chapter
- statistics and probability
- Topic
- total probability
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
2/5
The law of total probability decomposes an event over a partition of mutually exclusive cases, here the unseen first ball being red or blue, via P(B) = ΣP(B | Aᵢ)P(Aᵢ). This conditioning-on-the-first-draw technique is a recurring JEE Advanced device. The first ball is red with probability 4/10 and blue with probability 6/10. If the first ball was red, the urn has 3 red of 9 remaining, so the second is red with probability 3/9; if the first was blue, the urn has 4 red of 9, giving 4/9. Combining, P(second red) = (4/10)(3/9) + (6/10)(4/9) = (12 + 24)/90 = 36/90 = 2/5. Option 1/3 mistakenly conditions on only the red-first branch. Option 4/9 uses only the blue-first branch. Option 3/10 forgets to renormalise over the reduced count of nine. The result equals the original red fraction 4/10 = 2/5, illustrating the symmetry that without information, position does not change marginal probability. This is a deep and often surprising point: because the discarded first ball is never observed, the second draw is statistically indistinguishable from simply drawing a single ball from the original urn, so its red probability must coincide with the initial proportion. Any other answer would imply that merely relabelling which draw we call second could alter chances, which symmetry forbids. Plausibility check: 2/5 matches the proportion of red balls initially, exactly as expected when the discarded ball is unobserved, and it lies sensibly between the two branch probabilities 3/9 and 4/9, confirming the total-probability computation.
This hard difficulty mathematics question is from the chapter statistics and probability, covering the topic of total probability. It appeared in the 2025 exam.
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