Question

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

• A. $[0,11]$
• B. $[\frac{1}{3},\frac{2}{9}]$
• C. $[-\frac{1}{3},\frac{5}{9}]$
• D. $[-\frac{7}{9},\frac{4}{9}]$

Answer 1

The correct option is C $[-\frac{1}{3},\frac{5}{9}]$

$P\left(A\right)+P\left(B\right)\le 10\le P\left(A\right)\le 10\le P\left(B\right)\le 1\frac{3x+1}{3}+\frac{1-x}{4}\le 10\le \frac{3x+1}{3}\le 10\le \frac{1-x}{4}\le 1\frac{3x}{4}\le \frac{2}{3}-\frac{1}{4}0\le x+\frac{1}{3}\le 10\le 1-x\le 4\frac{3x}{4}\le \frac{8-3}{12}\frac{-1}{3}\le x\le \frac{2}{3}....\left(ii\right)-1\le -x\le 3or-3\le x\le \mathrm{1.......}\left(iii\right)\frac{3x}{4}\le \frac{5}{12}x\le \frac{5}{9}..........\left(i\right)from\left(i\right),\left(ii\right)and\left(iii\right)x\in \left[\frac{-1}{3},\frac{5}{9}\right]$

Option $C$ is correct answer.