Probability
Basic Probability
Grade 12
Question:
<p>If \(A\) and \(B\) are two events then</p><p>(a) \(P(A \cap B) \geq P(A) + P(B) - 1\)</p><p>(b) \(P(A \cap B) \geq P(A) + P(B)\)</p><p>(c) \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\)</p><p>(d) \(P(A \cap B) = P(A) + P(B) + P(A \cup B)\)</p>
<p>(a) \(P(A \cap B) \geq P(A) + P(B) - 1\)</p>
<p>(b) \(P(A \cap B) \geq P(A) + P(B)\)</p>
<p>(c) \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\)</p>
<p>(d) \(P(A \cap B) = P(A) + P(B) + P(A \cup B)\)</p>
Step-by-Step Solution
Key Concept: Use the fundamental identity P(A ∪ B) = P(A) + P(B) - P(A ∩ B) and the constraint that P(A ∪ B) ≤ 1 to derive bounds on P(A ∩ B). Rearranging gives P(A ∩ B) ≥ P(A) + P(B) - 1.
<p><strong>Step 1: Start with the fundamental probability identity</strong></p><p>For any two events A and B: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)</p><p><strong>Step 2: Check option (a)</strong></p><p>Rearranging the identity: P(A ∩ B) = P(A) + P(B) - P(A ∪ B)</p><p>Since P(A ∪ B) ≤ 1 (as a probability cannot exceed 1), we have:</p><p>P(A ∩ B) ≥ P(A) + P(B) - 1 ✓ <strong>TRUE</strong></p><p><strong>Step 3: Check option (b)</strong></p><p>P(A ∩ B) ≥ P(A) + P(B) would require P(A) + P(B) - P(A ∪ B) ≥ P(A) + P(B)</p><p>This means P(A ∪ B) ≤ 0, which is impossible. ✗ <strong>FALSE</strong></p><p><strong>Step 4: Check option (c)</strong></p><p>P(A ∩ B) = P(A) + P(B) - P(A ∪ B) is the direct rearrangement of the fundamental identity. ✓ <strong>TRUE</strong></p><p><strong>Step 5: Check option (d)</strong></p><p>The formula has a '+' sign instead of '-' before P(A ∪ B), which contradicts the probability axiom. ✗ <strong>FALSE</strong></p><p><strong>∴ Answer: A, C</strong></p>
Correct Answer: A,C