AP EAMCET · Maths · Probability
Events \(A, B\) and \(C\) are mutually exclusive events such that \(P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4}\) and \(P(C)=\frac{1-2 x}{2}\). The set of possible values of \(x\) are in the interval
- 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 \([0,1]\)
Answer & Solution
Correct Answer
(A) \(\left[\frac{1}{3}, \frac{1}{2}\right]\)
Step-by-step Solution
Detailed explanation
\(A, B, C\) are mutually exclusive. \[ \begin{aligned} & P(A \cap B \cap C)=0, P(A \cap B)=0, \\ & \quad P(B \cap C)=P(C \cap A)=0 \\ & \quad P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4}, P(C)=\frac{1-2 x}{2} \\ & \therefore P(A), P(B), P(C) \in[0,1] \end{aligned} \] Now,…
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