Events $$A, B, C$$  are mutually exclusive events such that $$P\left( A \right) = \frac{{3x + 1}}{3},P\left( B \right) = \frac{{1 - x}}{4}\,{\text{and }}P\left( C \right) = \frac{{1 - 2x}}{2}$$           The set of possible values of $$x$$ are in the interval.

Solution :
$$P\left( A \right) = \frac{{3x + 1}}{3},P\left( B \right) = \frac{{1 - x}}{4}\,{\text{and }}P\left( C \right) = \frac{{1 - 2x}}{2}$$
$$\because $$ For any event $$E,0 \leqslant P\left( E \right) \leqslant 1$$
$$\eqalign{ & \Rightarrow \,\,0 \leqslant \frac{{3x + 1}}{3} \leqslant 1,0 \leqslant \frac{{1 - x}}{4} \leqslant 1\,\,{\text{and }}0 \leqslant \frac{{1 - 2x}}{2} \leqslant 1 \cr & \Rightarrow \,\, - 1 \leqslant 3x \leqslant 2, - 3 \leqslant x \leqslant 1\,\,{\text{and }} - 1 \leqslant 2x \leqslant 1 \cr & \Rightarrow \,\, - \frac{1}{3} \leqslant x \leqslant \frac{2}{3} \leqslant - 3 \leqslant x \leqslant 1,\,\,{\text{and }} - \frac{1}{2} \leqslant x \leqslant \frac{1}{2} \cr} $$
Also for mutually exclusive events $$A, B, C,$$
$$\eqalign{ & P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) \cr & \Rightarrow \,\,P\left( {A \cup B \cup C} \right) = \frac{{3x + 1}}{3} + \frac{{1 - x}}{4} + \frac{{1 - 2x}}{2} \cr & \therefore \,\,0 \leqslant \frac{{1 + 3x}}{3} + \frac{{1 - x}}{4} + \frac{{1 - 2x}}{2} \leqslant 1 \cr & 0 \leqslant 13 - 3x \leqslant 12 \cr & \Rightarrow \,\,1 \leqslant 3x \leqslant 13 \cr & \Rightarrow \,\,\frac{1}{3} \leqslant x \leqslant \frac{{13}}{3} \cr} $$
Considering all inequations, we get
$$\eqalign{ & {\text{max}}\left\{ { - \frac{1}{3}, - 3, - \frac{1}{2},\frac{1}{3}} \right\} \leqslant x \leqslant {\text{min}}\left\{ {\frac{2}{3},1,\frac{1}{2},\frac{{13}}{3}} \right\} \cr & \frac{1}{3} \leqslant x \leqslant \frac{1}{2} \cr & \Rightarrow \,\,x \in \left[ {\frac{1}{3},\frac{1}{2}} \right] \cr} $$