Probability
Mutually Exclusive Events
Grade 12
Question:
<p>If the events \(A\) and \(B\) are mutually exclusive events such that \(P(A) = \dfrac{3x+1}{3}\) and \(P(B) = \dfrac{1-x}{4}\), then the set of possible real values of \(x\) lies in the interval</p>
<p>(1) \([0, 1]\)</p>
<p>(2) \(\left[-\dfrac{1}{3}, \dfrac{5}{9}\right]\)</p>
<p>(3) \(\left[-\dfrac{7}{9}, \dfrac{4}{9}\right]\)</p>
<p>(4) \(\left[\dfrac{1}{3}, \dfrac{2}{3}\right]\)</p>
Step-by-Step Solution
Key Concept: For mutually exclusive events, we need P(A) ≥ 0, P(B) ≥ 0, and P(A) + P(B) ≤ 1 (since they don't overlap). These three constraints must be satisfied simultaneously to find the valid range of x.
<p><strong>Step 1:</strong> Apply the constraint P(A) ≥ 0</p><p>$$\frac{3x+1}{3} \geq 0 \implies 3x + 1 \geq 0 \implies x \geq -\frac{1}{3}$$</p><p><strong>Step 2:</strong> Apply the constraint P(B) ≥ 0</p><p>$$\frac{1-x}{4} \geq 0 \implies 1 - x \geq 0 \implies x \leq 1$$</p><p><strong>Step 3:</strong> Apply the constraint P(A) ≤ 1</p><p>$$\frac{3x+1}{3} \leq 1 \implies 3x + 1 \leq 3 \implies x \leq \frac{2}{3}$$</p><p><strong>Step 4:</strong> Apply the constraint P(B) ≤ 1 (automatically satisfied when x ≥ -1/3)</p><p><strong>Step 5:</strong> For mutually exclusive events, P(A) + P(B) ≤ 1</p><p>$$\frac{3x+1}{3} + \frac{1-x}{4} \leq 1$$</p><p>$$\frac{4(3x+1) + 3(1-x)}{12} \leq 1$$</p><p>$$\frac{12x + 4 + 3 - 3x}{12} \leq 1$$</p><p>$$\frac{9x + 7}{12} \leq 1 \implies 9x + 7 \leq 12 \implies x \leq \frac{5}{9}$$</p><p><strong>Step 6:</strong> Combine all constraints: $x \geq -\frac{1}{3}$, $x \leq \frac{5}{9}$, $x \leq \frac{2}{3}$</p><p>The binding upper limit is $x \leq \frac{5}{9}$</p><p>∴ Answer: $x \in \left[-\frac{1}{3}, \frac{5}{9}\right]$</p>
Correct Answer: D