If the events A and B are mutually exclusive events such that $$P(A) = \frac{3x+1}{3}$$ and $$P(B) = \frac{1-x}{4}$$, then the set of possible values of x lies in the interval :

We are given that events A and B are mutually exclusive, meaning they cannot occur simultaneously. Therefore, the probability of their intersection is zero: $$P(A \cap B) = 0$$. Additionally, since they are mutually exclusive, the probability of their union is the sum of their individual probabilities: $$P(A \cup B) = P(A) + P(B)$$.

The probabilities are given as:

$$P(A) = \frac{3x+1}{3}$$

$$P(B) = \frac{1-x}{4}$$

As probabilities, they must satisfy the following conditions:

1. $$0 \leq P(A) \leq 1$$
2. $$0 \leq P(B) \leq 1$$
3. $$0 \leq P(A \cup B) \leq 1$$, which simplifies to $$0 \leq P(A) + P(B) \leq 1$$ because $$P(A \cup B) = P(A) + P(B)$$.

Since probabilities are non-negative, the condition $$P(A) + P(B) \geq 0$$ is automatically satisfied when both $$P(A) \geq 0$$ and $$P(B) \geq 0$$. Thus, we focus on the upper bound: $$P(A) + P(B) \leq 1$$.

We now derive the constraints step by step.

Constraint from $$P(A) \geq 0$$:

$$\frac{3x+1}{3} \geq 0$$

Multiply both sides by 3 (positive, so inequality direction remains the same):

$$3x + 1 \geq 0$$

$$3x \geq -1$$

$$x \geq -\frac{1}{3}$$

Constraint from $$P(A) \leq 1$$:

$$\frac{3x+1}{3} \leq 1$$

Multiply both sides by 3:

$$3x + 1 \leq 3$$

$$3x \leq 2$$

$$x \leq \frac{2}{3}$$

Constraint from $$P(B) \geq 0$$:

$$\frac{1-x}{4} \geq 0$$

Multiply both sides by 4:

$$1 - x \geq 0$$

$$x \leq 1$$

Constraint from $$P(B) \leq 1$$:

$$\frac{1-x}{4} \leq 1$$

Multiply both sides by 4:

$$1 - x \leq 4$$

$$-x \leq 3$$

Multiply both sides by -1 (reverse inequality direction):

$$x \geq -3$$

Constraint from $$P(A) + P(B) \leq 1$$:

$$\frac{3x+1}{3} + \frac{1-x}{4} \leq 1$$

Find a common denominator (12):

$$\frac{4(3x+1) + 3(1-x)}{12} \leq 1$$

Simplify the numerator:

$$4(3x+1) = 12x + 4$$

$$3(1-x) = 3 - 3x$$

So,

$$12x + 4 + 3 - 3x = 9x + 7$$

Thus,

$$\frac{9x + 7}{12} \leq 1$$

Multiply both sides by 12:

$$9x + 7 \leq 12$$

$$9x \leq 5$$

$$x \leq \frac{5}{9}$$

Now, we combine all constraints:

• From $$P(A) \geq 0$$: $$x \geq -\frac{1}{3}$$
• From $$P(A) \leq 1$$: $$x \leq \frac{2}{3}$$
• From $$P(B) \geq 0$$: $$x \leq 1$$
• From $$P(B) \leq 1$$: $$x \geq -3$$
• From $$P(A) + P(B) \leq 1$$: $$x \leq \frac{5}{9}$$

The most restrictive lower bound is the maximum of the lower limits: $$\max\left(-\frac{1}{3}, -3\right) = -\frac{1}{3}$$ (since $$-\frac{1}{3} > -3$$).

The most restrictive upper bound is the minimum of the upper limits: $$\min\left(\frac{2}{3}, 1, \frac{5}{9}\right)$$. Since $$\frac{5}{9} \approx 0.555$$, $$\frac{2}{3} \approx 0.666$$, and 1, the smallest is $$\frac{5}{9}$$.

Therefore, the combined constraints are:

$$-\frac{1}{3} \leq x \leq \frac{5}{9}$$

We verify the endpoints:

• At $$x = -\frac{1}{3}$$:

  $$P(A) = \frac{3\left(-\frac{1}{3}\right) + 1}{3} = \frac{-1 + 1}{3} = 0$$

  $$P(B) = \frac{1 - \left(-\frac{1}{3}\right)}{4} = \frac{1 + \frac{1}{3}}{4} = \frac{\frac{4}{3}}{4} = \frac{1}{3}$$

  $$P(A) + P(B) = 0 + \frac{1}{3} = \frac{1}{3} \leq 1$$

  All probabilities are valid.

• At $$x = \frac{5}{9}$$:

  $$P(A) = \frac{3\left(\frac{5}{9}\right) + 1}{3} = \frac{\frac{15}{9} + \frac{9}{9}}{3} = \frac{\frac{24}{9}}{3} = \frac{\frac{8}{3}}{3} = \frac{8}{9}$$

  $$P(B) = \frac{1 - \frac{5}{9}}{4} = \frac{\frac{4}{9}}{4} = \frac{4}{36} = \frac{1}{9}$$

  $$P(A) + P(B) = \frac{8}{9} + \frac{1}{9} = 1 \leq 1$$

  All probabilities are valid.

Values outside this interval violate the constraints. For example:

• If $$x < -\frac{1}{3}$$, say $$x = -0.4$$, then $$P(A) = \frac{3(-0.4) + 1}{3} = \frac{-1.2 + 1}{3} = \frac{-0.2}{3} < 0$$, invalid.
• If $$x > \frac{5}{9}$$, say $$x = 0.6$$, then $$P(A) = \frac{3(0.6) + 1}{3} = \frac{1.8 + 1}{3} = \frac{2.8}{3} \approx 0.933$$, $$P(B) = \frac{1 - 0.6}{4} = 0.1$$, and $$P(A) + P(B) \approx 1.033 > 1$$, invalid.

Thus, the set of possible values of x is the interval $$\left[-\frac{1}{3}, \frac{5}{9}\right]$$.

Comparing with the options:

A. [0, 1]

B. $$\left[\frac{1}{3}, \frac{2}{3}\right]$$

C. $$\left[-\frac{1}{3}, \frac{5}{9}\right]$$

D. $$\left[-\frac{7}{9}, \frac{4}{9}\right]$$

The interval $$\left[-\frac{1}{3}, \frac{5}{9}\right]$$ matches option C.

Hence, the correct answer is Option C.