A) \[\left[ \frac{1}{3},\frac{1}{2} \right]\]
B) \[\left[ \frac{1}{3},\frac{2}{3} \right]\]
C) \[\left[ \frac{1}{3},\frac{13}{3} \right]\]
D) \[[0,\,1]\]
Correct Answer: A
Solution :
We know that, \[0\le P\,(A\cup B\cup C)\le 1\] \[\Rightarrow \] \[0\le P(A)+P(B)+P(C)\le 1\] [\[\because \] A, B, C are mutually exclusive events] \[\Rightarrow \] \[0\le \left( \frac{3x+1}{3} \right)+\left( \frac{1-x}{4} \right)+\left( \frac{1-2x}{2} \right)\le 1\] \[\Rightarrow \] \[0\le \frac{12x+4+3-3x+6-12x}{12}\le 1\] \[\Rightarrow \] \[0\le 13-3x\le 12\] \[\Rightarrow \] \[-13\le -3x\le -1\] \[\Rightarrow \] \[13\ge 3x\ge 1\] \[\Rightarrow \] \[\frac{1}{3}\le x\le \frac{13}{3}\] Also, \[0\le P(A)\le 1,0\le P(B)\le 1,0\le P(C)\le 1\] \[\Rightarrow \]\[0\le \frac{3x+1}{3}\le 1,\,0\le \frac{1-x}{4}\le 1,0\le \frac{1-2x}{2}\le 1\] \[\Rightarrow \] \[-1\le x\le \frac{2}{3},-3\le x\le 1,-1\le x\le \frac{1}{2}\] \[\therefore \] From Eqs. (i) and (ii), \[x\in \left[ \frac{1}{3},\frac{1}{2} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
