If `(1 +3p )/( 3 ) , (1-p )/(4) , (1- 2p )/( 2)` are probabilities of three mutually exclusive events then

Options

A. `p= 0, 1/2`
B. `p=3/4,0`
C. `p=1/3, 1/2`
D. none of these

Answer

To solve the problem, we need to find the range of the variable \( p \) such that the expressions given represent valid probabilities for mutually exclusive events.

### Step-by-Step Solution:

1. **Identify the probabilities:**
The probabilities given are:
\[ P_1 = \frac{1 + 3p}{3}, \quad P_2 = \frac{1 - p}{4}, \quad P_3 = \frac{1 - 2p}{2} \]

2. **Set up inequalities for each probability:**
Since probabilities must be between 0 and 1, we can set up the following inequalities for each probability.

- For \( P_1 \):
\[ 0 \leq \frac{1 + 3p}{3} \leq 1 \]

- For \( P_2 \):
\[ 0 \leq \frac{1 - p}{4} \leq 1 \]

- For \( P_3 \):
\[ 0 \leq \frac{1 - 2p}{2} \leq 1 \]

3. **Solve the inequalities:**

- **For \( P_1 \)**:
\[ 0 \leq 1 + 3p \implies 3p \geq -1 \implies p \geq -\frac{1}{3} \]
\[ 1 + 3p \leq 3 \implies 3p \leq 2 \implies p \leq \frac{2}{3} \]
Thus, from \( P_1 \):
\[ -\frac{1}{3} \leq p \leq \frac{2}{3} \]

- **For \( P_2 \)**:
\[ 0 \leq 1 - p \implies p \leq 1 \]
\[ 1 - p \leq 4 \implies -p \leq 3 \implies p \geq -3 \]
Thus, from \( P_2 \):
\[ -3 \leq p \leq 1 \]

- **For \( P_3 \)**:
\[ 0 \leq 1 - 2p \implies 2p \leq 1 \implies p \leq \frac{1}{2} \]
\[ 1 - 2p \leq 2 \implies -2p \leq 1 \implies p \geq -\frac{1}{2} \]
Thus, from \( P_3 \):
\[ -\frac{1}{2} \leq p \leq \frac{1}{2} \]

4. **Combine the results:**
Now, we combine the results from all three inequalities:
- From \( P_1 \): \( -\frac{1}{3} \leq p \leq \frac{2}{3} \)
- From \( P_2 \): \( -3 \leq p \leq 1 \)
- From \( P_3 \): \( -\frac{1}{2} \leq p \leq \frac{1}{2} \)

The overlapping range is:
\[ -\frac{1}{3} \leq p \leq \frac{1}{2} \]

### Final Answer:
The value of \( p \) lies between \( -\frac{1}{3} \) and \( \frac{1}{2} \).