Events $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive events such that…
Events $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive events such that $\mathrm{P}(\mathrm{A})=\frac{3 \mathrm{x}+1}{3}, \mathrm{P}(\mathrm{B})=\frac{\mathrm{x}-1}{4}$ and $\mathrm{P}(\mathrm{C})=\frac{1-2 \mathrm{x}}{4}$. The set of possible values of $\mathrm{x}$ are in the interval.
$[0,1]$
$\left[\frac{1}{3}, \frac{1}{2}\right]$
$\left[\frac{1}{3}, \frac{2}{3}\right]$
$\left[\frac{1}{3}, \frac{13}{3}\right]$
Solution
$\mathrm{P}(\mathrm{A})=\frac{3 \mathrm{x}+1}{3}, \mathrm{P}(\mathrm{B})=\frac{1-\mathrm{x}}{4}, \mathrm{P}(\mathrm{C})=\frac{1-2 \mathrm{x}}{2}$
These are mutually exclusive
$0 \leq \frac{3 x+1}{3} \leq 1, \quad 0 \leq \frac{1-x}{4} \leq 1$ and $0 \leq \frac{1-2 x}{2} \leq 1$
$-1 \leq 3 x \leq 2,-3 \leq x \leq 1$ and $-1 \leq 2 x \leq 1$
$-\frac{1}{3} \leq x \leq \frac{2}{3},-3 \leq x \leq 1$, and $-\frac{1}{2} \leq x \leq \frac{1}{2}$
Also $0 \leq \frac{1+3 x}{3}+\frac{1-x}{4}+\frac{1-2 x}{2} \leq 1$
$0 \leq 13-3 x \leq 12 \Rightarrow 1 \leq 3 x \leq 13 \Rightarrow \frac{1}{3} \leq x \leq \frac{13}{3}$
$\max \left\{-\frac{1}{3},-3,-\frac{1}{2}, \frac{1}{3}\right\} \leq \mathrm{x} \leq \min \left\{\frac{2}{3}, 1, \frac{1}{2}, \frac{13}{3}\right\}$
$\frac{1}{3} \leq x \leq \frac{1}{2} \Rightarrow x \in\left[\frac{1}{3}, \frac{1}{2}\right]$