If  $\frac{1+3p}{3},\frac{1-2p}{2}$ are the probabilities of two mutually exclusive events then$P\in$

$0\le \frac{1+3p}{3}\le 1$      $0\le \frac{1-2p}{2}\le 1$

$0\le 1+3p\le 3$         $0\le 1-2p\le 2$

$-1\le 3p\le 2$             $-1\le -2p\le 1$

$\frac{-1}{3}\le p\le \frac{2}{3}$           $\frac{-1}{2}\le p\le \frac{1}{2}$

From these two:$\frac{-1}{3}\le p\le \frac{1}{2}$

$p\in \left[\frac{-1}{3},\frac{1}{2}\right]$