Question

Events $A,B,C$ are mutually exclusive events such that $P\left(A\right)=\frac{3x+1}{3},$ and $P\left(B\right)=\frac{1-x}{4},$ and $P\left(C\right)=\frac{1-2x}{2}.$ The set of possible values of $x$ is in the interval

• A. $[\frac{1}{3},\frac{2}{3}]$
• B. $[\frac{1}{3},\frac{13}{3}]$
• C. $[0,1]$
• D. $[\frac{1}{3},\frac{1}{2}]$

Answer 1

The correct option is D $[\frac{1}{3},\frac{1}{2}]$

$P\left(A\right)=\frac{3x+1}{3},P\left(B\right)=\frac{1-x}{4},P\left(C\right)=\frac{1-2x}{2}\because 0\le P\le 1\Rightarrow 0\le \frac{3x+1}{3}\le 1\Rightarrow 0\le 3x+1\le 3\Rightarrow \frac{-1}{3}\le x\le \frac{2}{3}similarly0\le \frac{1-x}{4}\le 1\Rightarrow -4\le x-1\le 0\Rightarrow -3\le x\le 1similarly0\le \frac{1-2x}{2}\le 1\Rightarrow -2\le 2x-1\le 0\Rightarrow \frac{-1}{2}\le x\le \frac{1}{2}Also0\le P(A\cup B\cup C)\le 1\Rightarrow 0\le P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)\le 1\Rightarrow 0\le P\left(A\right)+P\left(B\right)+P\left(C\right)\le 1$
Since A , B and C are mutually exhaustive hence $P(A\cap B)=P(B\cap C)=P(C\cap A)=P(A\cap B\cap C)=0$
$\Rightarrow 0\le \frac{3x+1}{3}+\frac{1-x}{4}+\frac{1-2x}{2}\le 1\Rightarrow 0\le \frac{13-3x}{12}\le 1\Rightarrow -12\le 3x-13\le 0\Rightarrow \frac{1}{3}\le x\le \frac{13}{3}$
comparing all the inequalities of x and taking the intersection we get
$\frac{1}{3}\le x\le \frac{1}{2}orx\in [\frac{1}{3},\frac{1}{2}]$