If `(1-3p)/(2),(1+4p)/(3) and (1+p)/(6)` are the probabilities of three mutually exclusive events, then the set of all values of p is_____

Options

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

Answer

To solve the problem, we need to find the set of values of \( p \) for which the expressions \( \frac{1-3p}{2} \), \( \frac{1+4p}{3} \), and \( \frac{1+p}{6} \) are valid probabilities of mutually exclusive events. This means that each probability must be between 0 and 1, and the sum of the probabilities must also be less than or equal to 1.

### Step 1: Set up the inequalities for each probability

1. For the first probability:
\[ 0 \leq \frac{1-3p}{2} \leq 1 \]

2. For the second probability:
\[ 0 \leq \frac{1+4p}{3} \leq 1 \]

3. For the third probability:
\[ 0 \leq \frac{1+p}{6} \leq 1 \]

### Step 2: Solve the inequalities

**Inequality 1:**
\[ 0 \leq \frac{1-3p}{2} \leq 1 \]
- From \( \frac{1-3p}{2} \geq 0 \):
\[ 1 - 3p \geq 0 \implies 3p \leq 1 \implies p \leq \frac{1}{3} \]

- From \( \frac{1-3p}{2} \leq 1 \):
\[ 1 - 3p \leq 2 \implies -3p \leq 1 \implies p \geq -\frac{1}{3} \]

Thus, from the first inequality, we have:
\[ -\frac{1}{3} \leq p \leq \frac{1}{3} \]

**Inequality 2:**
\[ 0 \leq \frac{1+4p}{3} \leq 1 \]
- From \( \frac{1+4p}{3} \geq 0 \):
\[ 1 + 4p \geq 0 \implies 4p \geq -1 \implies p \geq -\frac{1}{4} \]

- From \( \frac{1+4p}{3} \leq 1 \):
\[ 1 + 4p \leq 3 \implies 4p \leq 2 \implies p \leq \frac{1}{2} \]

Thus, from the second inequality, we have:
\[ -\frac{1}{4} \leq p \leq \frac{1}{2} \]

**Inequality 3:**
\[ 0 \leq \frac{1+p}{6} \leq 1 \]
- From \( \frac{1+p}{6} \geq 0 \):
\[ 1 + p \geq 0 \implies p \geq -1 \]

- From \( \frac{1+p}{6} \leq 1 \):
\[ 1 + p \leq 6 \implies p \leq 5 \]

Thus, from the third inequality, we have:
\[ -1 \leq p \leq 5 \]

### Step 3: Combine the inequalities

Now we need to find the intersection of the three sets of inequalities:
1. From the first inequality: \( -\frac{1}{3} \leq p \leq \frac{1}{3} \)
2. From the second inequality: \( -\frac{1}{4} \leq p \leq \frac{1}{2} \)
3. From the third inequality: \( -1 \leq p \leq 5 \)

The intersection of these inequalities is:
\[ -\frac{1}{4} \leq p \leq \frac{1}{3} \]

### Final Answer

Thus, the set of all values of \( p \) is:
\[ p \in \left[-\frac{1}{4}, \frac{1}{3}\right] \]