Q87 / 155 · WBJEE 2022 · Mathematics

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

  • A
    $[0,1]$
  • Correct answer
    B
    $\left[\frac{1}{3}, \frac{1}{2}\right]$
  • C
    $\left[\frac{1}{3}, \frac{2}{3}\right]$
  • D
    $\left[\frac{1}{3}, \frac{13}{3}\right]$

Solution

$0 \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C}) \leq 1$ and $0 \leq \mathrm{P}(\mathrm{A}) \leq 1,0 \leq \mathrm{P}(\mathrm{B}) \leq 1,0 \leq \mathrm{P}(\mathrm{C}) \leq 1 \Rightarrow \frac{1}{3} \leq \mathrm{x} \leq \frac{1}{2}$

Hence, the correct answer is option: 'B'.

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