Question

A, B, C are mutually exclusive events such that $P(A) = \frac{3x + 1}{3}$, $P(B) = \frac{1 - x}{4}$, and $P(C) = \frac{1 - 2x}{2}$. Then the set of possible values of $x$ are in

• A. [0, 1]
• B. [1/3, 1/2]
• C. [1/3, 2/3]
• D. [1/3, 13/3]

Answer 1

The Correct Option is B

Given: \( A \), \( B \), and \( C \) are mutually exclusive events, and the probabilities of these events are given by: - \( P(A) = \frac{3x + 1}{3} \) - \( P(B) = \frac{1 - x}{4} \) - \( P(C) = \frac{1 - 2x}{2} \) We are tasked with finding the set of possible values of \( x \) such that these probabilities are valid (i.e., they are between 0 and 1). Step 1: Conditions for valid probabilities. For any probability \( P \), it must satisfy: $$ 0 \leq P \leq 1 $$ Thus, we will apply this condition to each of the given probabilities. Step 2: Apply the conditions to each probability. 1. For \( P(A) = \frac{3x + 1}{3} \): $$ 0 \leq \frac{3x + 1}{3} \leq 1 $$ Multiply through by 3: $$ 0 \leq 3x + 1 \leq 3 $$ Subtract 1 from all parts: $$ -1 \leq 3x \leq 2 $$ Divide by 3: $$ -\frac{1}{3} \leq x \leq \frac{2}{3} $$ So, the valid range for \( x \) from this inequality is \( x \in \left[ -\frac{1}{3}, \frac{2}{3} \right] \). 2. For \( P(B) = \frac{1 - x}{4} \): $$ 0 \leq \frac{1 - x}{4} \leq 1 $$ Multiply through by 4: $$ 0 \leq 1 - x \leq 4 $$ Subtract 1 from all parts: $$ -1 \leq -x \leq 3 $$ Multiply by -1 (which reverses the inequality signs): $$ 1 \geq x \geq -3 $$ So, the valid range for \( x \) from this inequality is \( x \in \left[ -3, 1 \right] \). 3. For \( P(C) = \frac{1 - 2x}{2} \): $$ 0 \leq \frac{1 - 2x}{2} \leq 1 $$ Multiply through by 2: $$ 0 \leq 1 - 2x \leq 2 $$ Subtract 1 from all parts: $$ -1 \leq -2x \leq 1 $$ Multiply by -1 (which reverses the inequality signs): $$ 1 \geq 2x \geq -1 $$ Divide by 2: $$ \frac{1}{2} \geq x \geq -\frac{1}{2} $$ So, the valid range for \( x \) from this inequality is \( x \in \left[ -\frac{1}{2}, \frac{1}{2} \right] \). Step 3: Find the intersection of the three ranges. From the three conditions, the valid range for \( x \) must satisfy all three inequalities. The ranges are: - \( x \in \left[ -\frac{1}{3}, \frac{2}{3} \right] \) - \( x \in \left[ -3, 1 \right] \) - \( x \in \left[ -\frac{1}{2}, \frac{1}{2} \right] \) The intersection of these three ranges is: $$ x \in \left[ \frac{1}{3}, \frac{1}{2} \right] $$ Conclusion: The set of possible values of \( x \) is \( \left[ \frac{1}{3}, \frac{1}{2} \right] \). Thus, the correct answer is: [1/3, 1/2]