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>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{1}{5}\)</p>
Step-by-Step Solution
Key Concept: Use De Morgan's law: P(A' ∩ B') = P((A ∪ B)') to find P(A ∪ B), then apply conditional probability formula P(B|(A ∪ B')) = P(B ∩ (A ∪ B'))/P(A ∪ B'). Simplify B ∩ (A ∪ B') = B ∩ A since B ⊆ (A ∪ B').
<p><strong>Step 1:</strong> Find P(A ∪ B) using De Morgan's law.</p><p>P(A' ∩ B') = 0.5 ⟹ P((A ∪ B)') = 0.5</p><p>∴ P(A ∪ B) = 1 - 0.5 = 0.5</p><p><strong>Step 2:</strong> Find P(A ∪ B').</p><p>P(A ∪ B') = P(A) + P(B') - P(A ∩ B')</p><p>P(B') = 1 - 0.4 = 0.6</p><p>P(A ∩ B') = P(A) - P(A ∩ B) = P(A) - [P(A) + P(B) - P(A ∪ B)]</p><p>P(A ∩ B) = 0.3 + 0.4 - 0.5 = 0.2</p><p>P(A ∩ B') = 0.3 - 0.2 = 0.1</p><p>P(A ∪ B') = 0.3 + 0.6 - 0.1 = 0.8</p><p><strong>Step 3:</strong> Find P(B ∩ (A ∪ B')).</p><p>B ∩ (A ∪ B') = (B ∩ A) ∪ (B ∩ B') = (B ∩ A) = A ∩ B</p><p>P(B ∩ (A ∪ B')) = P(A ∩ B) = 0.2</p><p><strong>Step 4:</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>∴ Answer: <strong>1/4 or 0.25</strong></p>
Correct Answer: C