Key Concepts and Formulas

• Axiom of Probability: For any event $E$, its probability $P(E)$ must satisfy $0 \le P(E) \le 1$. This means probabilities cannot be negative or greater than 1.
• Mutually Exclusive Events: If events $A, B, C$ are mutually exclusive, it means that no two of them can occur simultaneously. In this case, the probability of their union is the sum of their individual probabilities: $P(A \cup B \cup C) = P(A) + P(B) + P(C)$
• Probability of Union: The probability of the union of any events must also satisfy the axiom of probability, i.e., $0 \le P(A \cup B \cup C) \le 1$.

Step-by-Step Solution

We are given the probabilities of three mutually exclusive events $A, B, C$ in terms of $x$: $P(A) = \frac{3x + 1}{3}$ $P(B) = \frac{1 - x}{4}$ $P(C) = \frac{1 - 2x}{2}$

Our goal is to find the set of possible values of $x$ that satisfy all the fundamental rules of probability.

Step 1: Apply Probability Axiom to Individual Events Each individual probability $P(A)$, $P(B)$, and $P(C)$ must be between 0 and 1, inclusive.

• For $P(A)$: $0 \le \frac{3x + 1}{3} \le 1$ Multiplying by 3: $0 \le 3x + 1 \le 3$ Subtracting 1: $-1 \le 3x \le 2$ Dividing by 3: $-\frac{1}{3} \le x \le \frac{2}{3}$ So, $x \in \left[ -\frac{1}{3}, \frac{2}{3} \right]$.

• For $P(B)$: $0 \le \frac{1 - x}{4} \le 1$ Multiplying by 4: $0 \le 1 - x \le 4$ Subtracting 1: $-1 \le -x \le 3$ Multiplying by -1 (and reversing inequalities): $-3 \le x \le 1$ So, $x \in \left[ -3, 1 \right]$.

• For $P(C)$: $0 \le \frac{1 - 2x}{2} \le 1$ Multiplying by 2: $0 \le 1 - 2x \le 2$ Subtracting 1: $-1 \le -2x \le 1$ Dividing by -2 (and reversing inequalities): $-\frac{1}{2} \le x \le \frac{1}{2}$ So, $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$.

Step 2: Find the Intersection of Individual Probability Intervals To satisfy all individual probability conditions, $x$ must be in the intersection of the intervals found in Step 1: $x \in \left[ -\frac{1}{3}, \frac{2}{3} \right] \cap \left[ -3, 1 \right] \cap \left[ -\frac{1}{2}, \frac{1}{2} \right]$ The most restrictive lower bound is $\max\left( -\frac{1}{3}, -3, -\frac{1}{2} \right) = -\frac{1}{3}$. The most restrictive upper bound is $\min\left( \frac{2}{3}, 1, \frac{1}{2} \right) = \frac{1}{2}$. Thus, from individual probabilities, $x \in \left[ -\frac{1}{3}, \frac{1}{2} \right]$.

Step 3: Apply Probability Axiom to the Union of Mutually Exclusive Events Since $A, B, C$ are mutually exclusive, $P(A \cup B \cup C) = P(A) + P(B) + P(C)$. The probability of their union must also be between 0 and 1.

First, calculate the sum $P(A) + P(B) + P(C)$: $P(A) + P(B) + P(C) = \frac{3x + 1}{3} + \frac{1 - x}{4} + \frac{1 - 2x}{2}$ To add these fractions, find a common denominator, which is 12: $= \frac{4(3x + 1)}{12} + \frac{3(1 - x)}{12} + \frac{6(1 - 2x)}{12}$ $= \frac{(12x + 4) + (3 - 3x) + (6 - 12x)}{12}$ $= \frac{12x - 3x - 12x + 4 + 3 + 6}{12}$ $= \frac{-3x + 13}{12}$

Now, apply the axiom $0 \le P(A \cup B \cup C) \le 1$: $0 \le \frac{-3x + 13}{12} \le 1$ Multiplying by 12: $0 \le -3x + 13 \le 12$ Subtracting 13: $-13 \le -3x \le -1$ Dividing by -3 (and reversing inequalities): $\frac{-1}{{-3}} \le x \le \frac{-13}{{-3}}$ $\frac{1}{3} \le x \le \frac{13}{3}$ So, $x \in \left[ \frac{1}{3}, \frac{13}{3} \right]$.

Step 4: Combine All Constraints The set of possible values for $x$ must satisfy all conditions derived from individual probabilities and the sum of probabilities. We need to find the intersection of the interval from Step 2 and the interval from Step 3: $x \in \left[ -\frac{1}{3}, \frac{1}{2} \right] \cap \left[ \frac{1}{3}, \frac{13}{3} \right]$ The lower bound is $\max\left( -\frac{1}{3}, \frac{1}{3} \right) = \frac{1}{3}$. The upper bound is $\min\left( \frac{1}{2}, \frac{13}{3} \right) = \frac{1}{2}$ (since $\frac{1}{2} = 0.5$ and $\frac{13}{3} \approx 4.33$). Therefore, the set of possible values for $x$ is $x \in \left[ \frac{1}{3}, \frac{1}{2} \right]$.

Common Mistakes & Tips

• Forgetting $P(E) \le 1$: Students often remember $P(E) \ge 0$ but forget that probability cannot exceed 1. Both bounds are crucial.
• Ignoring $P(\text{Union}) \le 1$: For mutually exclusive events, their sum must also adhere to the $0 \le P(E) \le 1$ rule. This often provides the most restrictive bounds for $x$.
• Incorrectly Handling Inequalities: When multiplying or dividing an inequality by a negative number, remember to reverse the direction of the inequality signs.

Summary

To find the set of possible values for $x$, we applied the fundamental axioms of probability. First, we ensured that each individual probability $P(A)$, $P(B)$, and $P(C)$ was between 0 and 1. This yielded an initial range for $x$. Next, since the events are mutually exclusive, we calculated the sum of their probabilities, $P(A) + P(B) + P(C)$, and constrained this sum to also be between 0 and 1. This provided further restrictions on $x$. The final set of possible values for $x$ is the intersection of all these derived intervals. The derived interval is $\left[ \frac{1}{3}, \frac{1}{2} \right]$.

Final Answer

The final answer is $\boxed{\left[ {{1 \over 3},{1 \over 2}} \right]}$, which corresponds to option (B).