If $\frac{1+4p}{4},\frac{1-p}{2}$ and $\frac{1-2p}{2}$ are the probabilities of three mutually exclusive events, then value of p is 

$\frac{1+4p}{4},\frac{1-p}{2},\frac{1-2p}{2}$ are probabilities of the three mutually exclusive events, then

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

and $0\le \frac{1+4\mathrm{p}}{4}+\frac{1-\mathrm{p}}{2}+\frac{1-2\mathrm{p}}{2}\le 1$

$\begin{array}{l}\therefore -\frac{1}{4}\le p\le \frac{3}{4},-1\le p\le 1,-\frac{1}{2}\le p\le \frac{1}{2},\frac{1}{2}\le p\le \frac{5}{2}\\ \therefore \frac{1}{2}\le p\le \frac{1}{2}\end{array}$

[The intersection of above four intervals] 

$\therefore \mathrm{p}=\frac{1}{2}$