<p>A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability that the lot contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelfth testing?</p>
<p>(a) \(\frac{99}{190}\)</p>
<p>(b) \(\frac{89}{1900}\)</p>
<p>(c) \(\frac{99}{1900}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The testing ends at draw 12 when all defective articles are found by the 12th draw, meaning the 12th article is the last defective one.
<p><strong>Step 1:</strong> For testing to end at the 12th draw, the 12th article must be defective and the first 11 articles must contain all remaining defective articles except one.</p><p><strong>Step 2:</strong> Case 1: Exactly 2 defective articles (probability 0.4). We need exactly 1 defective in first 11 draws, and the 12th is the second defective. Ways = C(2,1) × C(18,10) / C(20,11)</p><p><strong>Step 3:</strong> Case 2: Exactly 3 defective articles (probability 0.6). We need exactly 2 defective in first 11 draws, and the 12th is the third defective. Ways = C(3,2) × C(17,9) / C(20,11)</p><p><strong>Step 4:</strong> P(ends at 12th) = 0.4 × [C(2,1) × C(18,10) / C(20,11)] + 0.6 × [C(3,2) × C(17,9) / C(20,11)]</p><p>= 0.4 × (2 × 43758 / 167960) + 0.6 × (3 × 24310 / 167960) = 0.4 × (87516/167960) + 0.6 × (72930/167960) ≈ 99/1900</p><p>∴ Answer is (c) $\frac{99}{1900}$</p>
Correct Answer: C