Probability
Independent Events
Grade 12
Question:
<p>A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events \(A\), \(B\) and \(C\) are defined as follows:<br>\(A = \)(first bulb is defective)<br>\(B = \)(second bulb is non-defective)<br>\(C = \)(two bulbs are both defective or both non-defective)<br>Then</p>
<p>\(A\) and \(B\) are independent</p>
<p>\(B\) and \(C\) are independent</p>
<p>\(A\) and \(C\) are independent</p>
<p>\(A\), \(B\) and \(C\) are pairwise independent</p>
Step-by-Step Solution
Key Concept: With replacement, each draw is independent. Check pairwise independence (P(A∩B)=P(A)P(B), etc.) AND mutual independence (P(A∩B∩C)=P(A)P(B)P(C)) separately, as pairwise independence doesn't guarantee mutual independence.
<p><strong>Step 1: Calculate individual probabilities</strong></p><p>P(A) = P(first defective) = 50/100 = 1/2</p><p>P(B) = P(second non-defective) = 50/100 = 1/2</p><p>P(C) = P(both same type) = P(both defective) + P(both non-defective) = (1/2)(1/2) + (1/2)(1/2) = 1/2</p><p><strong>Step 2: Check pairwise independence</strong></p><p>P(A∩B) = P(first defective AND second non-defective) = (1/2)(1/2) = 1/4 = P(A)P(B) ✓</p><p>P(A∩C) = P(first defective AND both same) = P(both defective) = 1/4 = P(A)P(C) ✓</p><p>P(B∩C) = P(second non-defective AND both same) = P(both non-defective) = 1/4 = P(B)P(C) ✓</p><p><strong>Step 3: Check mutual independence</strong></p><p>P(A∩B∩C) = P(first defective AND second non-defective AND both same) = 0 (impossible: first defective but both same type requires second defective)</p><p>P(A)P(B)P(C) = (1/2)(1/2)(1/2) = 1/8 ≠ 0</p><p><strong>Conclusion:</strong> A, B, C are <strong>pairwise independent but NOT mutually independent</strong>.</p><p>∴ Answer: D</p>
Correct Answer: D