Question

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

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

Answer 1

Answer: (B)

$P\left(A\right)=\frac{3x+1}{3},P\left(B\right)=\frac{1-x}{4},$
$P\left(C\right)=\frac{1-2x}{2}$
$\because$ For any event $E,0\le P\left(E\right)\le 1$
$\Rightarrow 0\le \frac{3x+1}{3}\le 1,0\le \frac{1-x}{4}\le 1$
and $0\le \frac{1-2x}{2}\le 1$
$\Rightarrow -1\le 3x\le 2,-3\le x\le 1$ and $-1\le 2x\le 1$
$\Rightarrow -\frac{1}{3}\le x\le \frac{2}{3}\le -3\le x\le 1,$
and $-\frac{1}{2}\le x\le \frac{1}{2}$
Also for mutually exclusive events A, B, C,
$P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)$
$\Rightarrow P\left(A\cup B\cup C\right)=\frac{3x+1}{3}+\frac{1-x}{4}+\frac{1-2x}{2}$
$\therefore 0\le \frac{1+3x}{3}+\frac{1-x}{4}+\frac{1-2x}{2}\le 1$
$0\le 13-3x\le 12\Rightarrow 1\le 3x\le 13$
$\Rightarrow \frac{1}{3}\le x\le \frac{13}{3}$
Considering all inequations, we get
$\max \left\{-\frac{1}{3},-3,-\frac{1}{2},\frac{1}{3}\right\}\le x\le \min \left\{\frac{2}{3},1,\frac{1}{2},\frac{13}{3}\right\}$
$\frac{1}{3}\le x\le \frac{1}{2}\Rightarrow x\in \left[\frac{1}{3},\frac{1}{2}\right]$