Mathematics - Probability Question with Solution | TestHub

$\mathrm{P}(\mathrm{A})=\frac{3\mathrm{x}+1}{3},\mathrm{P}(\mathrm{B})=\frac{1-\mathrm{x}}{4},\mathrm{P}(\mathrm{C})=\frac{1-2\mathrm{x}}{2}$ These are mutually exclusive $0\le \frac{3x+1}{3}\le 1,0\le \frac{1-x}{4}\le 1$ and $0\le \frac{1-2x}{2}\le 1$ $-1\le 3x\le 2,-3\le x\le 1$ and $-1\le 2x\le 1$ $-\frac{1}{3}\le x\le \frac{2}{3},-3\le x\le 1$, and $-\frac{1}{2}\le x\le \frac{1}{2}$ Also $0\le \frac{1+3x}{3}+\frac{1-x}{4}+\frac{1-2x}{2}\le 1$ $0\le 13-3x\le 12\Rightarrow 1\le 3x\le 13\Rightarrow \frac{1}{3}\le x\le \frac{13}{3}$ $\max \left\{-\frac{1}{3},-3,-\frac{1}{2},\frac{1}{3}\right\}\le \mathrm{x}\le \min \left\{\frac{2}{3},1,\frac{1}{2},\frac{13}{3}\right\}$ $\frac{1}{3}\le x\le \frac{1}{2}\Rightarrow x\in \left[\frac{1}{3},\frac{1}{2}\right]$