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Step-by-Step Solution
Step 1: Write down the given probabilities
We have events A, B, and C with:
$P(A) = \frac{3x + 1}{3}$,
$P(B) = \frac{1 - x}{4}$, and
$P(C) = \frac{1 - 2x}{2}.$
Step 2: Apply the basic probability constraints
For any event X, $0 \le P(X) \le 1$. Thus:
1) For $P(A)$:
$$
0 \le \frac{3x + 1}{3} \le 1
\quad\Longrightarrow\quad 0 \le 3x + 1 \le 3,
$$
which simplifies to
$$
-1 \le 3x \le 2
\quad\Longrightarrow\quad -\frac{1}{3} \le x \le \frac{2}{3}.
$$
2) For $P(B)$:
$$
0 \le \frac{1 - x}{4} \le 1
\quad\Longrightarrow\quad 0 \le 1 - x \le 4.
$$
This yields
$$
1 - x \ge 0 \quad\Longrightarrow\quad x \le 1,
$$
and
$$
1 - x \le 4 \quad\Longrightarrow\quad x \ge -3.
$$
So together,
$$
-3 \le x \le 1.
$$
3) For $P(C)$:
$$
0 \le \frac{1 - 2x}{2} \le 1
\quad\Longrightarrow\quad 0 \le 1 - 2x \le 2.
$$
This yields
$$
1 - 2x \ge 0 \quad\Longrightarrow\quad 2x \le 1 \quad\Longrightarrow\quad x \le \frac{1}{2},
$$
and
$$
1 - 2x \le 2 \quad\Longrightarrow\quad -2x \le 1
\quad\Longrightarrow\quad x \ge -\frac{1}{2}.
$$
So together,
$$
-\frac{1}{2} \le x \le \frac{1}{2}.
$$
Step 3: Use the mutually exclusive property
Since A, B, and C are mutually exclusive, the total probability of their union is:
$$
P(A \cup B \cup C) = P(A) + P(B) + P(C).
$$
We must also have
$$
0 \le P(A) + P(B) + P(C) \le 1.
$$
Step 4: Compute $P(A) + P(B) + P(C)$ and set bounds
Calculate:
$$
P(A) + P(B) + P(C)
= \frac{3x + 1}{3} + \frac{1 - x}{4} + \frac{1 - 2x}{2}.
$$
Simplify this expression step by step (common denominator is 12):
$$
\frac{3x + 1}{3} = \frac{4(3x + 1)}{12} = \frac{12x + 4}{12},
$$
$$
\frac{1 - x}{4} = \frac{3(1 - x)}{12} = \frac{3 - 3x}{12},
$$
$$
\frac{1 - 2x}{2} = \frac{6(1 - 2x)}{12} = \frac{6 - 12x}{12}.
$$
Adding these gives:
$$
\frac{12x + 4}{12} + \frac{3 - 3x}{12} + \frac{6 - 12x}{12}
= \frac{12x + 4 + 3 - 3x + 6 - 12x}{12}
= \frac{(12x - 3x - 12x) + (4 + 3 + 6)}{12}.
$$
$$
= \frac{-3x + 13}{12}.
$$
Hence,
$$
P(A \cup B \cup C) = \frac{13 - 3x}{12}.
$$
We need
$$
0 \le \frac{13 - 3x}{12} \le 1.
$$
From $0 \le \frac{13 - 3x}{12}$, we get
$$
13 - 3x \ge 0 \quad\Longrightarrow\quad 3x \le 13 \quad\Longrightarrow\quad x \le \frac{13}{3}.
$$
From $\frac{13 - 3x}{12} \le 1$, we get
$$
13 - 3x \le 12 \quad\Longrightarrow\quad 3x \ge 1 \quad\Longrightarrow\quad x \ge \frac{1}{3}.
$$
So from the total probability constraint,
$$
\frac{1}{3} \le x \le \frac{13}{3}.
$$
Step 5: Combine all the interval constraints
From the individual constraints:
1) $-\frac{1}{3} \le x \le \frac{2}{3},$
2) $-3 \le x \le 1,$
3) $-\frac{1}{2} \le x \le \frac{1}{2},$
and from the union’s constraint:
4) $\frac{1}{3} \le x \le \frac{13}{3}.$
We look for the intersection of these intervals:
• The largest lower bound among
$-\frac{1}{3}, -3, -\frac{1}{2}, \frac{1}{3}$
is
$\frac{1}{3}.$
• The smallest upper bound among
$\frac{2}{3}, 1, \frac{1}{2}, \frac{13}{3}$
is
$\frac{1}{2}.$
Therefore, the valid range of $x$ is:
$$
\frac{1}{3} \le x \le \frac{1}{2}.
$$
Final Answer
The set of possible values of $x$ is
$$
\left[ \frac{1}{3}, \frac{1}{2} \right].
$$
