Let A, B and C be three mutually exclusive events such that P(A) = `p_(1), P(B) = p_(2) " and " P( C) = p_(3)`. Then,
Let `p_(1) = 1/3(1+3p), p_(2) = 1/4(1-p) " and " p_(3) = 1/4(1-2p)`, then p belongs to

Options

A. (-1, 0)
B. `(-2/3,2/3)`
C. `(-1/3,1)`
D. `(-1/9,1/4)`

Answer

To solve the problem, we need to find the values of \( p \) such that the probabilities \( P(A) \), \( P(B) \), and \( P(C) \) are valid and satisfy the conditions of mutual exclusivity.

### Step 1: Write down the probabilities
Given:
- \( P(A) = p_1 = \frac{1}{3}(1 + 3p) \)
- \( P(B) = p_2 = \frac{1}{4}(1 - p) \)
- \( P(C) = p_3 = \frac{1}{4}(1 - 2p) \)

### Step 2: Set up the inequality for the sum of probabilities
Since \( A \), \( B \), and \( C \) are mutually exclusive events, we have:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) \leq 1 \]
This leads to:
\[ p_1 + p_2 + p_3 \leq 1 \]

### Step 3: Substitute the values of \( p_1 \), \( p_2 \), and \( p_3 \)
Substituting the expressions for \( p_1 \), \( p_2 \), and \( p_3 \):
\[ \frac{1}{3}(1 + 3p) + \frac{1}{4}(1 - p) + \frac{1}{4}(1 - 2p) \leq 1 \]

### Step 4: Combine the terms
First, simplify the left-hand side:
\[ \frac{1}{3}(1 + 3p) + \frac{1}{4}(1 - p) + \frac{1}{4}(1 - 2p) = \frac{1 + 3p}{3} + \frac{1 - p + 1 - 2p}{4} \]
Combine the fractions:
\[ = \frac{1 + 3p}{3} + \frac{2 - 3p}{4} \]
Finding a common denominator (which is 12):
\[ = \frac{4(1 + 3p)}{12} + \frac{3(2 - 3p)}{12} = \frac{4 + 12p + 6 - 9p}{12} = \frac{10 + 3p}{12} \]

### Step 5: Set up the inequality
Now we have:
\[ \frac{10 + 3p}{12} \leq 1 \]
Multiply both sides by 12:
\[ 10 + 3p \leq 12 \]
Subtract 10 from both sides:
\[ 3p \leq 2 \]
Divide by 3:
\[ p \leq \frac{2}{3} \]

### Step 6: Find the lower bounds for \( p \)
Now, we also need to ensure that each probability is non-negative:
1. For \( P(A) \geq 0 \):
\[ \frac{1}{3}(1 + 3p) \geq 0 \implies 1 + 3p \geq 0 \implies p \geq -\frac{1}{3} \]
2. For \( P(B) \geq 0 \):
\[ \frac{1}{4}(1 - p) \geq 0 \implies 1 - p \geq 0 \implies p \leq 1 \]
3. For \( P(C) \geq 0 \):
\[ \frac{1}{4}(1 - 2p) \geq 0 \implies 1 - 2p \geq 0 \implies p \leq \frac{1}{2} \]

### Step 7: Combine the inequalities
From the inequalities we have:
- \( -\frac{1}{3} \leq p \leq \frac{2}{3} \)
- \( p \leq 1 \)
- \( p \leq \frac{1}{2} \)

The most restrictive bounds are:
\[ -\frac{1}{3} \leq p \leq \frac{1}{2} \]

### Final Answer
Thus, the values of \( p \) belong to the interval:
\[ p \in \left[-\frac{1}{3}, \frac{1}{2}\right] \]