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Probability
Bayes' Theorem, Conditional Probability
jee_main_2026_jan_21_shift_1
Grade None

Question:

A bag contains 10 balls out of which k are red and (10-k) are black, where 0 ≤ k ≤ 10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:
A. 7/11
B. 7/110
C. 14/55
D. 7/55

Step-by-Step Solution

Key Concept: Use Bayes' theorem with prior probabilities for k.
Step 1: P(k=1) = 1/11. P(3 black balls | k=1) = C(9,3)/C(10,3) = 84/120 = 7/10. Step 2: P(3 black balls | k) = C(10-k,3)/C(10,3). Sum over k=0 to 10: Σ C(10-k,3) = C(11,4) = 330. Step 3: P(k=1 | 3 black) = (1/11 × 7/10) / (1/11 × 330/120) = 14/55.
Correct Answer: C
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