Probability
Total Probability and Bayes' Theorem
Grade 12
Question:
<p>Three factories A, B and C produce the same product. The factory A produces twice as many as B produces while the factories B and C produce in the same quantity. It is known that 2% of the products of A as well as C are defective while 4% of the products of B are defective. All the products of the three factories are stocked together. If a product is selected at random from the stock, what is the probability that the product is defective?</p>
<p>\(\dfrac{1}{23}\)</p>
<p>\(\dfrac{1}{25}\)</p>
<p>\(\dfrac{1}{30}\)</p>
<p>\(\dfrac{2}{25}\)</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by finding the proportion each factory contributes to total production, then weight each factory's defect rate accordingly. The key is recognizing that A produces twice B's output, while B and C produce equally.
<p><strong>Step 1: Determine production ratios</strong></p><p>Let B produce x units. Then:</p><ul><li>Factory A produces: 2x units</li><li>Factory B produces: x units</li><li>Factory C produces: x units</li><li>Total production: 2x + x + x = 4x units</li></ul><p><strong>Step 2: Find proportion of each factory's output</strong></p><ul><li>P(from A) = 2x/4x = 1/2</li><li>P(from B) = x/4x = 1/4</li><li>P(from C) = x/4x = 1/4</li></ul><p><strong>Step 3: Identify defect rates</strong></p><ul><li>Defect rate of A: P(defective|A) = 2% = 0.02</li><li>Defect rate of B: P(defective|B) = 4% = 0.04</li><li>Defect rate of C: P(defective|C) = 2% = 0.02</li></ul><p><strong>Step 4: Apply total probability theorem</strong></p><p>P(defective) = P(defective|A)·P(A) + P(defective|B)·P(B) + P(defective|C)·P(C)</p><p>P(defective) = (0.02)(1/2) + (0.04)(1/4) + (0.02)(1/4)</p><p>P(defective) = 0.01 + 0.01 + 0.005 = 0.025 = 2.5%</p><p>∴ Answer: <strong>B</strong> (or 1/40 or 0.025)</p>
Correct Answer: B