Probability
Independent Events
Grade 12
Question:
<p>Event 'A' is independent of event \(B\), \(B \cup C\) and \(B \cap C\). If \(P(A) = \dfrac{1}{2}\), \(P(B) = \dfrac{1}{3}\) and \(P(C) = \dfrac{1}{4}\). Then:</p>
<p>\(P\left(\dfrac{A}{C}\right) = \dfrac{1}{2}\)</p>
<p>\(P\left(\dfrac{\overline{B \cup C}}{A}\right) = \dfrac{11}{12}\) (where \(B\) and \(C\) are independent events)</p>
<p>\(P\left(\dfrac{\overline{A}}{B \cap C}\right) = \dfrac{1}{2}\)</p>
<p>\(A\) and \(C\) are not independent events</p>
Step-by-Step Solution
Key Concept: If A is independent of B, B∪C, and B∩C simultaneously, then A must be independent of C as well. Use the independence condition P(A∩X) = P(A)·P(X) for each event, then verify which statements about conditional probabilities and joint probabilities hold true.
<p><strong>Step 1: Establish independence relationships</strong></p><p>Given: A is independent of B, (B∪C), and (B∩C)</p><p>From independence: P(A∩B) = P(A)·P(B) = (1/2)·(1/3) = 1/6</p><p>From independence with (B∩C): P(A∩B∩C) = P(A)·P(B∩C)</p><p><strong>Step 2: Determine independence with C</strong></p><p>Since A is independent of (B∪C): P(A∩(B∪C)) = P(A)·P(B∪C)</p><p>Using A∩(B∪C) = (A∩B)∪(A∩C):</p><p>P(A∩B) + P(A∩C) - P(A∩B∩C) = P(A)[P(B) + P(C) - P(B∩C)]</p><p>Substituting known independent values and solving shows A is independent of C</p><p><strong>Step 3: Verify candidate statements</strong></p><p><strong>Option A:</strong> P(A∩B∩C) = 1/24 ✓ (Using independence: (1/2)·(1/3)·(1/4) = 1/24)</p><p><strong>Option B:</strong> P(A|B∪C) = 1/2 ✓ (By independence of A with (B∪C))</p><p><strong>Option C:</strong> P(C|A∩B) = 1/4 ✓ (A independent of C means P(C|A∩B) = P(C) = 1/4)</p><p>∴ Answer: A, B, C</p>
Correct Answer: A,B,C