If $\frac{1+4p}{4}$, $\frac{1-p}{3}$ and $\frac{1-2p}{2}$ are the probabilities of three mutually exclusive events then $p$ may be:
Step-by-Step Solution
Key Concept: For mutually exclusive events, each probability must satisfy 0 ≤ P ≤ 1, and their sum must satisfy 0 ≤ P(A) + P(B) + P(C) ≤ 1. Apply all four constraints simultaneously: individual bounds on each expression plus the sum constraint.
From constraints $0 \leq \frac{1+4p}{4} \leq 1$, we get $-\frac{3}{4} \leq p \leq \frac{3}{4}$ ... (1). From $0 \leq \frac{1-p}{3} \leq 1$, we get $-2 \leq p \leq 1$ ... (2). From $0 \leq \frac{1-2p}{2} \leq 1$, we get $-\frac{1}{2} \leq p \leq \frac{1}{2}$ ... (3). Combined with $0 \leq P(A \cup B \cup C) \leq 1$ giving $\frac{1}{4} \leq p \leq \frac{13}{4}$ ... (4), the intersection yields $\frac{1}{4} \leq p \leq \frac{1}{2}$.
Correct Answer: 2,3