Probability
Conditional probability / Total probability
Grade 12
Question:
<p>Lot <em>A</em> consists of 5 good and 3 defective articles. Lot <em>B</em> consists of 3 good and 5 defective articles. A new lot <em>C</em> is formed by taking 3 articles from <em>A</em> and 4 articles from <em>B</em>. The probability that an article chosen at random from <em>C</em> 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 the probability that a randomly selected article from C is defective, considering it could come from either the 3 articles taken from A or the 4 articles taken from B.
<p><strong>Step 1:</strong> Find the probability that an article from lot A is defective.</p><p>P(defective from A) = 3/8</p><p><strong>Step 2:</strong> Find the probability that an article from lot B is defective.</p><p>P(defective from B) = 5/8</p><p><strong>Step 3:</strong> In lot C, there are 3 articles from A and 4 from B (total 7 articles). When choosing randomly from C, the probability of selecting from A is 3/7 and from B is 4/7.</p><p><strong>Step 4:</strong> Apply the 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>P(defective from C) = (3/7) × (3/8) + (4/7) × (5/8)</p><p>= 9/56 + 20/56 = 29/56</p><p>∴ Answer: C</p>
Correct Answer: C