If the events $\mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events such that…
If the events $\mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events such that $\mathrm{P}(\mathrm{A})=\frac{3 x+1}{3}$ and $\mathrm{P}(\mathrm{B})=\frac{1-x}{4}$, then the set of possible values of $x$ lies in the interval :
$[0,1]$
$\left[\frac{1}{3}, \frac{2}{3}\right]$
$\left[-\frac{1}{3}, \frac{5}{9}\right]$
$\left[-\frac{7}{9}, \frac{4}{9}\right]$
Solution
Since events A and B are mutually exclusive
$
\begin{aligned}
&\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=1 \\
&\Rightarrow \frac{3 x+1}{3}+\frac{1-x}{4}=1 \\
&\Rightarrow 12 x+4+3-3 x=12 \\
&\Rightarrow 9 x=5 \Rightarrow x=\frac{5}{9} \\
&\therefore x \in\left[-\frac{1}{3}, \frac{5}{9}\right]
\end{aligned}
$