Probability
Conditional Probability
Grade 12
Question:
<p>\(A\) and \(B\) are two events defined as follows:<br>\(A\): It rains today with \(P(A) = 40\%\)<br>\(B\): It rains tomorrow with \(P(B) = 50\%\)<br>Also, \(P\)(it rains today and tomorrow) \(= 30\%\)<br>Also, \(E_1: P((A \cap B)/(A \cup B))\) and \(E_2: P(\{(A \cap \bar{B}) \text{ or } (B \cup \bar{A})\}/(A \cup B))\). Then which of the following is/are true?</p>
<p>(1) \(A\) and \(B\) are independent</p>
<p>(2) \(P(A/B) < P(B/A)\)</p>
<p>(3) \(E_1\) and \(E_2\) are equiprobable</p>
<p>(4) \(P(A/(A \cup B)) = P(B/(A \cup B))\)</p>
Step-by-Step Solution
Key Concept: Use conditional probability formula P(X|Y) = P(X∩Y)/P(Y) and carefully parse the complex event descriptions. First compute P(A∪B) using inclusion-exclusion, then identify the exact events in numerator before applying the conditional probability definition.
<p><strong>Step 1: Identify given information</strong></p><p>P(A) = 0.4, P(B) = 0.5, P(A∩B) = 0.3</p><p><strong>Step 2: Calculate P(A∪B) using inclusion-exclusion</strong></p><p>P(A∪B) = P(A) + P(B) - P(A∩B) = 0.4 + 0.5 - 0.3 = 0.6</p><p><strong>Step 3: Calculate E₁ = P((A∩B)/(A∪B))</strong></p><p>E₁ = P(A∩B)/P(A∪B) = 0.3/0.6 = 1/2 = 0.5</p><p><strong>Step 4: Identify the event for E₂</strong></p><p>The event is: (A∩B̄) ∪ (B∩Ā) [symmetric difference - exactly one event occurs]</p><p>P((A∩B̄) ∪ (B∩Ā)) = P(A∩B̄) + P(B∩Ā) = [P(A) - P(A∩B)] + [P(B) - P(A∩B)]</p><p>= (0.4 - 0.3) + (0.5 - 0.3) = 0.1 + 0.2 = 0.3</p><p><strong>Step 5: Calculate E₂</strong></p><p>E₂ = P((A∩B̄) ∪ (B∩Ā))/(A∪B) = 0.3/0.6 = 1/2 = 0.5</p><p><strong>Step 6: Verify partition property</strong></p><p>Check: P(A∩B) + P(A∩B̄) + P(B∩Ā) = 0.3 + 0.1 + 0.2 = 0.6 = P(A∪B) ✓</p><p>Therefore: E₁ = 0.5 and E₂ = 0.5</p><p>∴ Both statements are true (Answer: 2, 3)</p>
Correct Answer: 2,3