Probability
Conditional Probability
Grade 12
Question:
<p>If two events \(A\) and \(B\) are such that \(P(A') = 0.3\), \(P(B) = 0.4\) and \(P(A \cap B') = 0.5\), then find the value of \(P[B/(A \cup B')]\).</p>
Step-by-Step Solution
Key Concept: Use the complement rule and conditional probability formula: P(B|A∪B') = P(B∩(A∪B'))/P(A∪B'). Since B and B' are mutually exclusive, B∩(A∪B') = B∩A, so we need P(A∩B) and P(A∪B').
<p><strong>Step 1:</strong> Extract given information</p><p>P(A') = 0.3 ⟹ P(A) = 0.7</p><p>P(B) = 0.4 ⟹ P(B') = 0.6</p><p>P(A∩B') = 0.5</p><p><strong>Step 2:</strong> Find P(A∩B)</p><p>Since A = (A∩B) ∪ (A∩B') and these are disjoint:</p><p>P(A) = P(A∩B) + P(A∩B')</p><p>0.7 = P(A∩B) + 0.5</p><p>P(A∩B) = 0.2</p><p><strong>Step 3:</strong> Simplify B∩(A∪B')</p><p>B∩(A∪B') = (B∩A) ∪ (B∩B') = B∩A (since B∩B' = ∅)</p><p>So P(B∩(A∪B')) = P(A∩B) = 0.2</p><p><strong>Step 4:</strong> Find P(A∪B')</p><p>P(A∪B') = P(A) + P(B') - P(A∩B')</p><p>P(A∪B') = 0.7 + 0.6 - 0.5 = 0.8</p><p><strong>Step 5:</strong> Apply conditional probability formula</p><p>P(B|A∪B') = P(B∩(A∪B'))/P(A∪B') = 0.2/0.8 = 1/4</p><p><strong>∴ Answer: 1/4</strong></p>
Correct Answer: 1/4