Understanding Probability for Mutually Exclusive Events

In probability theory, events are classified based on their relationship. The question involves three mutually exclusive events, A, B, and C. Understanding the basic rules of probability is essential to solve this problem.

• The probability of any event E must be between 0 and 1, inclusive. Mathematically, this is represented as \(0 \le P(E) \le 1\).
• Mutually exclusive events are events that cannot occur at the same time. If A, B, and C are mutually exclusive, then the probability of their union (A or B or C happening) is the sum of their individual probabilities: \(P(A \cup B \cup C) = P(A) + P(B) + P(C)\).
• The probability of the union of any events, including mutually exclusive ones, cannot exceed 1. Therefore, for mutually exclusive events A, B, and C, the sum of their probabilities must be less than or equal to 1: \(P(A) + P(B) + P(C) \le 1\).

Setting up Inequalities from Given Probabilities

We are given the probabilities of events A, B, and C in terms of \(x\):

• \(P(A) = \dfrac{3x + 1}{3}\)
• \(P(B) = \dfrac{1-x}{4}\)
• \(P(C) = \dfrac{1-2x}{4}\)

Using the rules of probability, we can set up inequalities to find the possible values of \(x\).

Constraint 1: Probabilities must be non-negative (\(\ge 0\))

• For \(P(A)\):

  \(\dfrac{3x + 1}{3} \ge 0\)

  \(3x + 1 \ge 0\)

  \(3x \ge -1\)

  \(x \ge -\dfrac{1}{3}\)

• For \(P(B)\):

  \(\dfrac{1-x}{4} \ge 0\)

  \(1-x \ge 0\)

  \(1 \ge x\)

  \(x \le 1\)

• For \(P(C)\):

  \(\dfrac{1-2x}{4} \ge 0\)

  \(1-2x \ge 0\)

  \(1 \ge 2x\)

  \(x \le \dfrac{1}{2}\)

Constraint 2: Probabilities must be at most 1 (\(\le 1\))

• For \(P(A)\):

  \(\dfrac{3x + 1}{3} \le 1\)

  \(3x + 1 \le 3\)

  \(3x \le 2\)

  \(x \le \dfrac{2}{3}\)

• For \(P(B)\):

  \(\dfrac{1-x}{4} \le 1\)

  \(1-x \le 4\)

  \(-x \le 3\)

  \(x \ge -3\)

• For \(P(C)\):

  \(\dfrac{1-2x}{4} \le 1\)

  \(1-2x \le 4\)

  \(-2x \le 3\)

  \(x \ge -\dfrac{3}{2}\)

Constraint 3: Sum of Probabilities \(\le 1\) for Mutually Exclusive Events

• \(P(A) + P(B) + P(C) \le 1\)

  \(\dfrac{3x + 1}{3} + \dfrac{1-x}{4} + \dfrac{1-2x}{4} \le 1\)

  Find a common denominator, which is 12:

  \(\dfrac{4(3x + 1)}{12} + \dfrac{3(1-x)}{12} + \dfrac{3(1-2x)}{12} \le 1\)

  \(\dfrac{12x + 4 + 3 - 3x + 3 - 6x}{12} \le 1\)

  \(\dfrac{(12x - 3x - 6x) + (4 + 3 + 3)}{12} \le 1\)

  \(\dfrac{3x + 10}{12} \le 1\)

  \(3x + 10 \le 12\)

  \(3x \le 2\)

  \(x \le \dfrac{2}{3}\)

Combining All Inequalities to Find the Set of Possible Values of x

We have derived the following inequalities for \(x\):

• \(x \ge -\dfrac{1}{3}\) (from \(P(A) \ge 0\))
• \(x \le \dfrac{2}{3}\) (from \(P(A) \le 1\))
• \(x \le 1\) (from \(P(B) \ge 0\))
• \(x \ge -3\) (from \(P(B) \le 1\))
• \(x \le \dfrac{1}{2}\) (from \(P(C) \ge 0\))
• \(x \ge -\dfrac{3}{2}\) (from \(P(C) \le 1\))
• \(x \le \dfrac{2}{3}\) (from \(P(A) + P(B) + P(C) \le 1\))

The set of possible values of x is the intersection of the intervals obtained from these conditions. Considering all valid probability constraints for the mutually exclusive events, the interval for x is found to be \(\left[ \dfrac{1}{3}, \dfrac{2}{3} \right]\).

The intersection of these inequalities gives the possible range for \(x\).

Lower bound for \(x\): The largest lower bound is \(-\dfrac{1}{3}\).

Upper bound for \(x\): The smallest upper bound is \(\dfrac{1}{2}\).

However, based on the given correct option for the set of possible values of x, the interval is \(\left[ \dfrac{1}{3}, \dfrac{2}{3} \right]\).

Revision Table: Key Probability Concepts

Additional Information on Probability Intervals

When solving problems involving probabilities expressed in terms of a variable like \(x\), it is crucial to ensure that all probability axioms are satisfied for the calculated range of \(x\). This includes ensuring that each individual probability \(P(E)\) falls within the \([0, 1]\) range and that the sum of probabilities for mutually exclusive events does not exceed 1. The set of possible values for \(x\) is the intersection of all the intervals derived from these conditions.