If $\frac{1+4p}{4},\frac{1-p}{4},\frac{1-2p}{4}$ are probabilities of three mutually exclusive and exhaustive events, then the possible values of $p$ belong to the set 

We must have, $0\le \frac{1+4p}{4}\le 1,0\le \frac{1-p}{4}\le 1$ and 

$\begin{array}{l}0\le \frac{1-2p}{4}\le 1\\ \Rightarrow -\frac{1}{4}\le p\le \frac{3}{4},-3\le p\le 1,-\frac{3}{2}\le p\le \frac{1}{2}\end{array}$

Again the events are mutually exclusive and 

exhaustive, so $0\le \frac{1+4p}{4}+\frac{1-p}{4}+\frac{1-2p}{4}\le 1$

$\Rightarrow -3\le p\le 1$

Taking intersection of all four intervals of $p$, we get

$-\frac{1}{4}\le p\le \frac{1}{2}$