Probability
Event
Grade 12
Question:
<p>Events A, B, C are mutually exclusive events such that \(P(A) = \dfrac{3x+1}{3}\), \(P(B) = \dfrac{1-x}{4}\) and \(P(C) = \dfrac{1-2x}{2}\). The set of possible values of \(x\) are in the interval</p>
<p>\(\left[\dfrac{1}{3}, \dfrac{1}{2}\right]\)</p>
<p>\(\left[\dfrac{1}{3}, \dfrac{2}{3}\right]\)</p>
<p>\(\left[\dfrac{1}{3}, \dfrac{13}{3}\right]\)</p>
<p>\([0, 1]\)</p>
Step-by-Step Solution
Key Concept: For mutually exclusive events, each probability must be non-negative and their sum cannot exceed 1. We need to find all values of x satisfying these constraints simultaneously.
<p><strong>Step 1: Set up constraints for non-negative probabilities</strong></p><p>Since A, B, C are events, each probability must be non-negative:</p><p>P(A) ≥ 0: $\frac{3x+1}{3} \geq 0 \Rightarrow 3x+1 \geq 0 \Rightarrow x \geq -\frac{1}{3}$</p><p>P(B) ≥ 0: $\frac{1-x}{4} \geq 0 \Rightarrow 1-x \geq 0 \Rightarrow x \leq 1$</p><p>P(C) ≥ 0: $\frac{1-2x}{2} \geq 0 \Rightarrow 1-2x \geq 0 \Rightarrow x \leq \frac{1}{2}$</p><p><strong>Step 2: Set up constraint for each probability ≤ 1</strong></p><p>P(A) ≤ 1: $\frac{3x+1}{3} \leq 1 \Rightarrow 3x+1 \leq 3 \Rightarrow x \leq \frac{2}{3}$</p><p>P(B) ≤ 1: $\frac{1-x}{4} \leq 1 \Rightarrow 1-x \leq 4$ (always true for reasonable x)</p><p>P(C) ≤ 1: $\frac{1-2x}{2} \leq 1 \Rightarrow 1-2x \leq 2$ (always true for x ≥ -\frac{1}{2}$)</p><p><strong>Step 3: Apply mutual exclusivity constraint</strong></p><p>Since A, B, C are mutually exclusive: $P(A) + P(B) + P(C) \leq 1$</p><p>$\frac{3x+1}{3} + \frac{1-x}{4} + \frac{1-2x}{2} \leq 1$</p><p>Multiply by 12: $4(3x+1) + 3(1-x) + 6(1-2x) \leq 12$</p><p>$12x + 4 + 3 - 3x + 6 - 12x \leq 12$</p><p>$-3x + 13 \leq 12$</p><p>$-3x \leq -1$</p><p>$x \geq \frac{1}{3}$</p><p><strong>Step 4: Combine all constraints</strong></p><p>From Step 1: $x \geq -\frac{1}{3}$, $x \leq 1$, $x \leq \frac{1}{2}$</p><p>From Step 2: $x \leq \frac{2}{3}$</p><p>From Step 3: $x \geq \frac{1}{3}$</p><p>Taking intersection: $\frac{1}{3} \leq x \leq \frac{1}{2}$</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A