If the events A and B arc mutually exclusive events such that $\mathrm{P}\left(\mathrm{A}\right)=\frac{3\mathrm{x}+1}{3}\text{ and }\mathrm{P}\left(\mathrm{B}\right)=\frac{1-\mathrm{x}}{4}$, then the set of possible real values of x lies in the interval

We must have,

$\begin{array}{l}0\le P\left(A\right)\le 1\\ \Rightarrow 0\le \frac{3x+1}{3}\le 1\\ \Rightarrow \frac{-1}{3}\le x\le \frac{2}{3} \left(1\right)\end{array}$

and $0\le \mathrm{P}\left(\mathrm{B}\right)\le 1$

$\Rightarrow 0\le \frac{1-\mathrm{x}}{4}\le 1$

$\Rightarrow -3\le \mathrm{x}\le 1$           (2)

and $0\le \mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)\le 1$

$\Rightarrow 0\le \frac{3\mathrm{x}+1}{3}+\frac{1-\mathrm{x}}{4}\le 1$

$\Rightarrow \frac{-7}{9}\le \mathrm{x}\le \frac{5}{9}$          (3)

From (1), (2) and (3), we get $\mathrm{x}\in \left[\frac{-1}{3},\frac{5}{9}\right]$.