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JEE Mains · Maths · STD 12 - 13. probability

If the events \(A\) and \(B\) are mutually exclusive events such that \(P\left( A \right) = \frac{{3x + 1}}{3}\) and \(P\left( B \right) = \frac{{1 - x}}{4}\), then the set of possible values of \(x\) lies in the interval

  1. A \([0, 1]\)
  2. B \(\left[ {\frac{1}{3},\frac{2}{3}} \right]\)
  3. C \(\left[ { - \frac{1}{3},\frac{5}{9}} \right]\)
  4. D \(\left[ { - \frac{7}{9},\frac{4}{9}} \right]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left[ { - \frac{1}{3},\frac{5}{9}} \right]\)

Step-by-step Solution

Detailed explanation

Since events \(A\) and \(B\) are mutually exclusive \(\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=1\) \(\Rightarrow \frac{3 x+1}{3}+\frac{1-x}{4}=1\) \(\Rightarrow 12 x+4+3-3 x=12\) \(\Rightarrow 9 x=5 \Rightarrow x=\frac{5}{9}\)…