If $(1+3p)/3,(1-p)/4$ and $(1-2p)/2$ are the probabilities of three mutually exclusive events, then the set of all values of $p$ is 

As $(1+3p)/3, (1-p)/4$ and $(1-2p)/2$ are the probabilities of three mutually exclusive events 

                                  $\frac{1+3p}{3}\ge 0,\frac{1-p}{4}\ge 0,\frac{1-2p}{2}\ge 0$

and                  $\frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{2}\le 1$

$\Rightarrow p\ge 1/3,p\le 1,p\le 1/2,p\ge 1/3.$

Thus,    $1/3\le p\le 1/2.$