Probability
Conditional Probability
Grade 12
Question:
<p>For any two events \(A\) and \(B\),</p>
<p>(a) \(P(A/B) = \frac{P(A \cap B) - P(B) - 1}{P(B)}\), \(P(B) \neq 0\)</p>
<p>(b) \(P(A \cap B) \neq P(A) - P(A \cap B)\)</p>
<p>(c) \(P(A \cap B) = 1 - P(A)P(B)\) if \(A, B\) are independent</p>
<p>(d) \(P(A \cap B) = 1 - P(A)P(B)\) if \(A, B\) are mutually exclusive</p>
Step-by-Step Solution
Key Concept: Use the fundamental probability axioms: P(A∪B) = P(A) + P(B) - P(A∩B) and the constraint that all probabilities must satisfy 0 ≤ P(E) ≤ 1 for any event E. The relationship between these must hold for ANY two events.
<p><strong>Step 1:</strong> Recall the addition rule for probability: P(A∪B) = P(A) + P(B) - P(A∩B)</p><p><strong>Step 2:</strong> Since P(A∪B) ≤ 1 (fundamental axiom), we have: P(A) + P(B) - P(A∩B) ≤ 1</p><p><strong>Step 3:</strong> Since P(A∩B) ≥ 0, the most restrictive case gives: P(A) + P(B) ≤ 1 + P(A∩B)</p><p><strong>Step 4:</strong> Also, P(A∩B) ≤ min{P(A), P(B)}, and since we need this for ANY events, the universal constraint is: <strong>P(A) + P(B) - 1 ≤ P(A∩B) ≤ min{P(A), P(B)}</strong></p><p><strong>Step 5:</strong> Equivalently, <strong>P(A∩B) ≥ P(A) + P(B) - 1</strong> and <strong>P(A∪B) ≤ P(A) + P(B)</strong></p><p>∴ Answer: C</p>
Correct Answer: C