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

Options

A. `[-(1)/(3),(1)/(3)]`
B. `[-(1)/(3),1]`
C. `[(1)/(13),1]`
D. `[(1)/(13),(1)/(3)]`

Answer

To solve the problem, we need to find the set of all values of \( p \) such that the given expressions represent valid probabilities for three mutually exclusive events. The probabilities given are:

1. \( P_1 = \frac{1 + 3p}{4} \)
2. \( P_2 = \frac{1 - p}{3} \)
3. \( P_3 = \frac{1 - 3p}{2} \)

### Step 1: Establish the conditions for probabilities

Since these are probabilities, they must satisfy the conditions:

\[ 0 \leq P_1 \leq 1 \]
\[ 0 \leq P_2 \leq 1 \]
\[ 0 \leq P_3 \leq 1 \]

### Step 2: Solve the inequalities for \( P_1 \)

For \( P_1 \):

\[ 0 \leq \frac{1 + 3p}{4} \leq 1 \]

**Lower Bound:**
\[ 1 + 3p \geq 0 \implies 3p \geq -1 \implies p \geq -\frac{1}{3} \]

**Upper Bound:**
\[ 1 + 3p \leq 4 \implies 3p \leq 3 \implies p \leq 1 \]

Thus, from \( P_1 \), we have:
\[ -\frac{1}{3} \leq p \leq 1 \]

### Step 3: Solve the inequalities for \( P_2 \)

For \( P_2 \):

\[ 0 \leq \frac{1 - p}{3} \leq 1 \]

**Lower Bound:**
\[ 1 - p \geq 0 \implies p \leq 1 \]

**Upper Bound:**
\[ 1 - p \leq 3 \implies -p \leq 2 \implies p \geq -2 \]

Thus, from \( P_2 \), we have:
\[ -2 \leq p \leq 1 \]

### Step 4: Solve the inequalities for \( P_3 \)

For \( P_3 \):

\[ 0 \leq \frac{1 - 3p}{2} \leq 1 \]

**Lower Bound:**
\[ 1 - 3p \geq 0 \implies 3p \leq 1 \implies p \leq \frac{1}{3} \]

**Upper Bound:**
\[ 1 - 3p \leq 2 \implies -3p \leq 1 \implies p \geq -\frac{1}{3} \]

Thus, from \( P_3 \), we have:
\[ -\frac{1}{3} \leq p \leq \frac{1}{3} \]

### Step 5: Combine the results

Now we combine the results from all three inequalities:

1. From \( P_1 \): \( -\frac{1}{3} \leq p \leq 1 \)
2. From \( P_2 \): \( -2 \leq p \leq 1 \)
3. From \( P_3 \): \( -\frac{1}{3} \leq p \leq \frac{1}{3} \)

The intersection of these intervals gives us:

\[ -\frac{1}{3} \leq p \leq \frac{1}{3} \]

### Conclusion

Thus, the set of all values of \( p \) such that the given expressions are valid probabilities is:

\[ \boxed{[-\frac{1}{3}, \frac{1}{3}]} \]