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: Multiple constraints from probability axioms must be simultaneously satisfied to determine the valid range of parameters.
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