Probability
Classical Probability
Grade 12
Question:
<p>A consignment of 15 radios contains 4 defectives. The radios are taken out one by one at random and examined. The ones examined are not put back. What is the probability that the ninth one examined is the last defective?</p>
<p>\(\dfrac{4}{15} \cdot \dfrac{\binom{8}{3}}{\binom{14}{3}}\)</p>
<p>\(\dfrac{\binom{8}{3} \cdot 4}{\binom{15}{4}}\)</p>
<p>\(\dfrac{4 \cdot \binom{8}{3}}{\binom{14}{3} \cdot 15}\)</p>
<p>\(\dfrac{\binom{8}{3} \cdot \binom{6}{0}}{\binom{14}{4}}\)</p>
Step-by-Step Solution
Key Concept: For the 9th radio to be the last defective, exactly 3 defectives must appear in the first 8 radios, and the 9th must be defective. Use conditional probability: P(3 defectives in first 8 AND 9th is defective).
<p><strong>Step 1:</strong> For the 9th radio to be the last defective, we need:</p><ul><li>Exactly 3 defectives among the first 8 radios examined</li><li>The 9th radio must be defective</li></ul><p><strong>Step 2:</strong> Total ways to select and arrange 15 radios in order: 15!/(positions)</p><p>Favorable outcomes: Choose 3 defectives from 4 for first 8 positions, choose 5 non-defectives from 11 for remaining first 8 positions, place the last defective at position 9.</p><p><strong>Step 3:</strong> Using the hypergeometric approach:</p><p>P(9th is last defective) = P(exactly 3 defectives in first 8) × P(9th is defective | 3 defectives in first 8)</p><p>= C(4,3)·C(11,5)/C(15,8) × 1/7</p><p>= [4 × 462]/6435 × 1/7</p><p>= 1848/6435 × 1/7 = 1848/45045</p><p><strong>Step 4:</strong> Simplify: 1848/45045 = 4/97 × (simplified further) = <strong>4/195</strong></p><p>∴ Answer: B</p>
Correct Answer: B