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

Answer

To solve the problem, we need to find the set of values of \( p \) such that the probabilities given by the expressions \( \frac{1 + 3p}{3} \), \( \frac{1 - p}{4} \), and \( \frac{1 - 2p}{2} \) are valid probabilities (i.e., they lie between 0 and 1) and also satisfy the condition of being mutually exclusive events.

### Step-by-Step Solution

1. **Set up the inequalities for each probability:**
- For the first probability:
\[ 0 \leq \frac{1 + 3p}{3} \leq 1 \]
- For the second probability:
\[ 0 \leq \frac{1 - p}{4} \leq 1 \]
- For the third probability:
\[ 0 \leq \frac{1 - 2p}{2} \leq 1 \]

2. **Solve the first inequality:**
- From \( 0 \leq \frac{1 + 3p}{3} \):
\[ 0 \leq 1 + 3p \implies 3p \geq -1 \implies p \geq -\frac{1}{3} \]
- From \( \frac{1 + 3p}{3} \leq 1 \):
\[ 1 + 3p \leq 3 \implies 3p \leq 2 \implies p \leq \frac{2}{3} \]
- Thus, from the first probability, we have:
\[ -\frac{1}{3} \leq p \leq \frac{2}{3} \]

3. **Solve the second inequality:**
- From \( 0 \leq \frac{1 - p}{4} \):
\[ 0 \leq 1 - p \implies p \leq 1 \]
- From \( \frac{1 - p}{4} \leq 1 \):
\[ 1 - p \leq 4 \implies -p \leq 3 \implies p \geq -3 \]
- Thus, from the second probability, we have:
\[ -3 \leq p \leq 1 \]

4. **Solve the third inequality:**
- From \( 0 \leq \frac{1 - 2p}{2} \):
\[ 0 \leq 1 - 2p \implies 2p \leq 1 \implies p \leq \frac{1}{2} \]
- From \( \frac{1 - 2p}{2} \leq 1 \):
\[ 1 - 2p \leq 2 \implies -2p \leq 1 \implies p \geq -\frac{1}{2} \]
- Thus, from the third probability, we have:
\[ -\frac{1}{2} \leq p \leq \frac{1}{2} \]

5. **Combine the inequalities:**
- From the first inequality: \( -\frac{1}{3} \leq p \leq \frac{2}{3} \)
- From the second inequality: \( -3 \leq p \leq 1 \)
- From the third inequality: \( -\frac{1}{2} \leq p \leq \frac{1}{2} \)

The intersection of these intervals gives:
\[ -\frac{1}{3} \leq p \leq \frac{1}{2} \]

6. **Check the mutual exclusivity condition:**
- The sum of the probabilities must be less than or equal to 1:
\[ \frac{1 + 3p}{3} + \frac{1 - p}{4} + \frac{1 - 2p}{2} < 1 \]
- Simplifying this gives:
\[ \frac{4 + 12p + 3 - 3p + 6 - 12p}{12} < 1 \]
\[ \frac{13 - 3p}{12} < 1 \implies 13 - 3p < 12 \implies 3p > 1 \implies p > \frac{1}{3} \]

7. **Final result:**
- Combining all conditions, we find:
\[ \frac{1}{3} < p \leq \frac{1}{2} \]

### Conclusion
The set of all values of \( p \) is:
\[ \left( \frac{1}{3}, \frac{1}{2} \right] \]