Probability
Mutually Exclusive Events / Probability Constraints
Grade 12
Question:
<p>Since \(A\), \(B\), \(C\) are mutually exclusive events with \(P(A) = \frac{3x+1}{3}\), \(P(B) = \frac{1-x}{4}\) and \(P(C) = \frac{1-2x}{2}\), the range of \(x\) such that all probabilities are valid is:</p>
<p>\(x \in \left[\frac{1}{3}, \frac{1}{2}\right]\)</p>
<p>\(x \in \left[\frac{1}{3}, \frac{13}{3}\right]\)</p>
<p>\(x \in \left[-\frac{1}{3}, \frac{2}{3}\right]\)</p>
<p>\(x \in \left[-3, 1\right]\)</p>
Step-by-Step Solution
Key Concept: For mutually exclusive events, each probability must satisfy 0 ≤ P ≤ 1, AND their sum must not exceed 1 (since P(A∪B∪C) ≤ 1). Apply all four constraints simultaneously to find the intersection of valid ranges.
<p><strong>Step 1: Apply constraint P(A) ∈ [0,1]</strong></p><p>0 ≤ (3x+1)/3 ≤ 1</p><p>0 ≤ 3x+1 ≤ 3</p><p>-1 ≤ 3x ≤ 2</p><p><strong>⟹ -1/3 ≤ x ≤ 2/3</strong></p><p><strong>Step 2: Apply constraint P(B) ∈ [0,1]</strong></p><p>0 ≤ (1-x)/4 ≤ 1</p><p>0 ≤ 1-x ≤ 4</p><p>-1 ≤ -x ≤ 3</p><p><strong>⟹ -3 ≤ x ≤ 1</strong></p><p><strong>Step 3: Apply constraint P(C) ∈ [0,1]</strong></p><p>0 ≤ (1-2x)/2 ≤ 1</p><p>0 ≤ 1-2x ≤ 2</p><p>-1 ≤ -2x ≤ 1</p><p><strong>⟹ -1/2 ≤ x ≤ 1/2</strong></p><p><strong>Step 4: Apply mutual exclusivity constraint P(A) + P(B) + P(C) ≤ 1</strong></p><p>(3x+1)/3 + (1-x)/4 + (1-2x)/2 ≤ 1</p><p>Multiply by 12: 4(3x+1) + 3(1-x) + 6(1-2x) ≤ 12</p><p>12x + 4 + 3 - 3x + 6 - 12x ≤ 12</p><p>-3x + 13 ≤ 12</p><p><strong>⟹ x ≥ 1/3</strong></p><p><strong>Step 5: Find intersection of all constraints</strong></p><p>From Steps 1-4: [-1/3, 2/3] ∩ [-3, 1] ∩ [-1/2, 1/2] ∩ [1/3, ∞)</p><p><strong>∴ Answer: x ∈ [1/3, 1/2]</strong></p>
Correct Answer: A