If `((1-3p))/2,((1+4p))/3,((1+p))/6` are the probabilities of three mutually excusing and exhaustive events, then the set of all values of `p` is a. `(0,1)` b. `(-1/4,1/3)` c. `(0,1/3)` d.

Explanation

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

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

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