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PROBABILITY — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSPROBABILITY1 Mark
MCQ
The events A, B, C are mutually exclusive events such that $\text{P(A)}=\frac{(3\text{x}+1)}{3,\text{P(B)}}=\frac{(\text{x}-1)}{4\text{ and}\text{ P}}\text{(C)}=\frac{(1-2\text{x})}{4}.$The set of possible values of x are in the interval:
  • ✓
    $\Big[\frac{1}{3},\frac{1}{2}\Big]$
  • B
    $\Big[\frac{1}{3},\frac{2}{3}\Big]$
  • C
    $\Big[\frac{1}{3},\frac{13}{3}\Big]$
  • D
    $\Big[0,1\Big]$
✓

Answer

Correct option: A.
$\Big[\frac{1}{3},\frac{1}{2}\Big]$
$\text{P(A)}=\frac{(3\text{x}+1)}{3}$
$\text{P(B)}=\frac{(\text{x}-1)}{4}$
$\text{P(C)}=\frac{(1-2\text{x})}{4}$
These are mutually exclusive events.
$\Rightarrow-1\leq3\text{x}\leq2,-3\leq\text{x}\leq,-\leq2\text{x}\leq1$
$\Rightarrow-\frac{1}{3}\leq\text{x}\leq\frac{2}{3},-2\leq\text{x}\leq1,-\frac{1}{2}\leq\text{x}\leq\frac{1}{2}$
Also, 0 $\leq\frac{(3\text{x}+1)}{3}+\frac{\text{x}-1}{4}+\frac{(1-2\text{x)}}{4\leq1}$
$\Rightarrow\frac{1}{3}\leq\text{x}\leq\frac{13}{3}$
$\Rightarrow\text{max}\Big\{\frac{-1}{3},-3,-\frac{1}{2},\frac{1}{3}\Big\}$
$\Rightarrow\frac{1}{3}\leq\text{x}\leq\frac{1}{2}$
$\Rightarrow\text{x}\in\Big[\frac{1}{3},\frac{1}{2}\Big]$

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