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Question: Answered & Verified by Expert
$A, B, C$ 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
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Options:
  • A $[0,1]$
  • 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:
1350 Upvotes Verified Answer
The correct answer is: $\left[\frac{1}{3}, \frac{1}{2}\right]$
$0 \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C}) \leq 1$ and $0 \leq \mathrm{P}(\mathrm{A}) \leq 1,0 \leq \mathrm{P}(\mathrm{B}) \leq 1,0 \leq \mathrm{P}(\mathrm{C}) \leq 1 \Rightarrow \frac{1}{3} \leq \mathrm{x} \leq \frac{1}{2}$

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