Question

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

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

Answer 1

Answer: (A)

$[\frac{1}{3},\frac{1}{2}]$

$A,B,C$ are mutually exclusive events then

$P\left(A\right)+P\left(B\right)+P\left(C\right)=1$;

and $0<P\left(A\right)<1$ ; $0<P\left(B\right)<1$ and $0<P\left(C\right)<1$

$0<\frac{3x+1}{3}<1\Rightarrow 3x+1>0\Rightarrow x>\frac{-1}{3}3x+1<3\Rightarrow 3x<2\Rightarrow x<\frac{2}{3}$

i.e., $xϵ(\frac{-1}{3},\frac{2}{3})$

$0<P\left(B\right)<1$

$\Rightarrow 0<\frac{x-1}{4}<1\Rightarrow 0<x-1<4\Rightarrow 1<x<5xϵ(1,5)0<P\left(C\right)<1\Rightarrow 0<\frac{1-2x}{2}<1\Rightarrow 0<1-2x<2-1<-2x<1\Rightarrow xϵ(\frac{-1}{2},\frac{1}{2})$

Also $P\left(A\right)+P\left(B\right)+P\left(C\right)=1$

$\frac{3x+1}{3}+\frac{x-1}{4}+\frac{1-2x}{2}=1\Rightarrow \frac{12x+4+3x-3+6-12x}{12}=1\Rightarrow 3x=12-7\Rightarrow x=\frac{5}{3}$

Taking all the intersection of all the cases the possible values of $x$ are in interval $[\frac{1}{3},\frac{1}{2}]$