Question

If events $A,B,C$ are mutually exclusive such that $P\left(A\right)=\frac{3x+1}{3},P\left(B\right)=\frac{1-x}{4},P\left(C\right)=\frac{1-2x}{2}$, then set of possible values of $x$ are 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 $[\frac{1}{3},\frac{1}{2}]$

Given $A,B,C$ are mutually exclusive
$\therefore 0\le P\left(A\right)+P\left(B\right)+P\left(C\right)\le 1.....\left(i\right)$
and $0\le P\left(A\right),P\left(B\right),P\left(C\right)\le \mathrm{1...}\left(ii\right)$
now on solving (i) and (ii) we get $\frac{1}{3}\le x\le \frac{1}{2}$.