Probability
Total Probability
Grade 12
Question:
<p>Lot \(A\) consists of 5 good and 3 defective articles. Lot \(B\) consists of 3 good and 5 defective articles. A new lot \(C\) is formed by taking 3 articles from \(A\) and 4 articles from \(B\). The probability that an article chosen at random from \(C\) is defective, is:</p>
<p>(a) \(\dfrac{1}{3}\)</p>
<p>(b) \(\dfrac{2}{5}\)</p>
<p>(c) \(\dfrac{29}{56}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by finding P(defective from A) and P(defective from B) separately, then weight them by their contributions to lot C (3/7 and 4/7 respectively).
<p><strong>Step 1:</strong> Find the composition of lot C.</p><p>From lot A (5 good, 3 defective): take 3 articles<br/>From lot B (3 good, 5 defective): take 4 articles<br/>Lot C contains: 3 + 4 = 7 articles total</p><p><strong>Step 2:</strong> Calculate expected number of defective articles from A in C.</p><p>P(article from A is defective) = 3/8<br/>Expected defective articles from A in C = 3 × (3/8) = 9/8</p><p><strong>Step 3:</strong> Calculate expected number of defective articles from B in C.</p><p>P(article from B is defective) = 5/8<br/>Expected defective articles from B in C = 4 × (5/8) = 20/8</p><p><strong>Step 4:</strong> Apply law of total probability.</p><p>P(defective from C) = P(from A) × P(defective|from A) + P(from B) × P(defective|from B)</p><p>= (3/7) × (3/8) + (4/7) × (5/8)</p><p>= 9/56 + 20/56</p><p>= 29/56</p><p><strong>∴ Answer: C (29/56)</strong></p>
Correct Answer: C