If $\frac{(1+3P)}{3},\frac{(1-p)}{4}$ and $\frac{(1 − 2p)}{2}$ are the probabilities of three mutually exclusive events, then the set of all values of p is

$\frac{1}{3}≤p ≤\frac{1}{2}$

Since $\frac{(1+3P)}{3}, \frac{(1-p)}{4} $ and $ (\frac{1-2p}{2})$ are the probabilities of the three events, we must have $0 ≤ \frac{1+3p}{3}≤1. 0 ≤ \frac{1-p}{4}≤1 $ and $ 0 ≤ \frac{1-2p}{2}≤ 1 ⇒ -1 ≤ 3p ≤ 2, -3 ≤ p ≤ 1 $ and $ -1 ≤ 2p ≤1 $

$⇒ -\frac{1}{3} ≤ p ≤ \frac{2}{3}, -3 ≤ p ≤ 1 $ and $ -\frac{1}{2} ≤ p ≤ \frac{1}{2}.$

Also as $\frac{1+3p}{3}, \frac{1-p}{4}$ and $\frac{1-2p}{2}$ are the probabilities of three mutually exclusive events, 

$0 ≤\frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{4}≤ 1 ⇒ 0 ≤4 + 12 p + 3 - 3p + 6 - 12 p ≤ 12 ⇒ \frac{1}{3} ≤ p ≤ \frac{13}{3}$

Thus the required values of p are such that   $ max\begin{Bmatrix} -\frac{1}{3}, -3, -\frac{1}{2}, \frac{1}{3}\end{Bmatrix}≤ p ≤ min \begin{Bmatrix}\frac{2}{3}, 1, \frac{1}{2}, \frac{13}{3}\end{Bmatrix} ⇒\frac{1}{3}≤p ≤\frac{1}{2}$