Probability
Bayes' Theorem
Grade 12
Question:
<p>The chances of defective screws in three boxes \(A\), \(B\), \(C\) are 1/5, 1/6, 1/7, respectively. A box is selected at random and a screw drawn from it at random is found to be defective. Then find the probability that it came from box \(A\).</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem: the probability that a screw came from box A given it's defective equals P(defective|A)×P(A) divided by the total probability of drawing a defective screw from any box.
<p><strong>Step 1: Identify given information</strong></p><p>P(A) = P(B) = P(C) = 1/3 (equal probability of selecting any box)</p><p>P(defective|A) = 1/5, P(defective|B) = 1/6, P(defective|C) = 1/7</p><p><strong>Step 2: Find total probability of drawing a defective screw using law of total probability</strong></p><p>P(defective) = P(defective|A)·P(A) + P(defective|B)·P(B) + P(defective|C)·P(C)</p><p>P(defective) = (1/5)·(1/3) + (1/6)·(1/3) + (1/7)·(1/3)</p><p>P(defective) = (1/3)[1/5 + 1/6 + 1/7]</p><p><strong>Step 3: Calculate the sum inside brackets</strong></p><p>LCM(5,6,7) = 210</p><p>1/5 + 1/6 + 1/7 = 42/210 + 35/210 + 30/210 = 107/210</p><p>P(defective) = (1/3)·(107/210) = 107/630</p><p><strong>Step 4: Apply Bayes' theorem</strong></p><p>P(A|defective) = [P(defective|A)·P(A)] / P(defective)</p><p>P(A|defective) = [(1/5)·(1/3)] / (107/630)</p><p>P(A|defective) = (1/15) / (107/630) = (1/15) × (630/107)</p><p>P(A|defective) = 630/(15×107) = 42/107</p><p>∴ <strong>Answer: 42/107</strong></p>
Correct Answer: 42/107