A) \[[-1/3,\,\,1/2]\]
B) \[[1/3,\,2/3]\]
C) \[[1/3,\,13/3]\]
D) \[[0,1]\]
E) None of these
Correct Answer: E
Solution :
Given, \[P(A)=\frac{3x+1}{3},\] \[P(B)=\frac{1-x}{4}\] and \[P(C)=\frac{1-2x}{2}\] Since, A,B and C are mutually exclusive events. \[\therefore \] \[0\le \frac{3x+1}{3}\le 1,\,0\le \frac{1-x}{4}\le 1,\,0\le \frac{1-2x}{2}\le 1\] and \[0\le \frac{3x+1}{3}+\frac{1-x}{4}+\frac{1-2x}{2}\le 1\] \[\Rightarrow \] \[0\le 3x+1\le 3,0\le 1-x\le 4,0\le 1-2x\le 2\] \[\Rightarrow \] \[-1\le 3x\le 2,-1\le -x\le 3,-1\le -2x\le 1\] \[\Rightarrow \] \[\frac{-1}{3}\le x\le \frac{2}{3},-3\le x\le 1,-\frac{1}{2}\le x\le \frac{1}{2}\] ?.(i) and \[0\le \frac{4(3x+1)+3(1-x)+6(1-2x)}{12}\le 1\] \[\Rightarrow \] \[0\le -3x+13\le 12\] \[\Rightarrow \] \[-13\le -3x\le -1\] \[\frac{1}{3}\le x\le \frac{13}{3}\] ?.(ii) \[\therefore \] From Eq (i) and (ii), we get \[\frac{1}{3}\le \frac{1}{2}\] \[\therefore \] \[x\in \left[ \frac{1}{3},\frac{1}{2} \right]\]
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