Question:
If the events A and B are mutually exclusive events such that \(P(A)=\frac{3x+1}{3}\) and \(P(B)=\frac{1-x}{4}\), then the set of possible values of \(x\) lies in the interval :
If the events A and B are mutually exclusive events such that \(P(A)=\frac{3x+1}{3}\) and \(P(B)=\frac{1-x}{4}\), then the set of possible values of \(x\) lies in the interval :
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Each probability lies in [0,1] and for mutually exclusive events \(P(A)+P(B)\le1\).
Updated On: Nov 4, 2026
- \([0,1]\)
- \(\left[\frac13,\frac23\right]\)
- \(\left[-\frac13,\frac59\right]\)
- \(\left[-\frac79,\frac49\right]\)
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The Correct Option is C
Solution and Explanation
\(0\le\frac{3x+1}{3}\le1\Rightarrow-\frac13\le x\le\frac23\).
\(0\le\frac{1-x}{4}\le1\Rightarrow-3\le x\le1\).
Mutually exclusive: \(P(A)+P(B)\le1\Rightarrow\frac{3x+1}{3}+\frac{1-x}{4}\le1\Rightarrow4(3x+1)+3(1-x)\le12\Rightarrow9x\le5\Rightarrow x\le\frac59\).
Intersection: \(-\frac13\le x\le\frac59\).
Answer: (3) \(\left[-\frac13,\frac59\right]\)
\(0\le\frac{1-x}{4}\le1\Rightarrow-3\le x\le1\).
Mutually exclusive: \(P(A)+P(B)\le1\Rightarrow\frac{3x+1}{3}+\frac{1-x}{4}\le1\Rightarrow4(3x+1)+3(1-x)\le12\Rightarrow9x\le5\Rightarrow x\le\frac59\).
Intersection: \(-\frac13\le x\le\frac59\).
Answer: (3) \(\left[-\frac13,\frac59\right]\)
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