COMEDK · Maths · 36. Probability
If the events A and B are mutually exclusive events such that \(P(A)=\dfrac{1}{3}(3 x+1)\) and \(P(B)=\dfrac{1}{4}(1-x)\) then the possible values of \(x\) lies in the interval
- A \(\left[-\dfrac{7}{9}, \dfrac{4}{9}\right]\)
- B \(\left[-\dfrac{1}{3}, \dfrac{5}{9}\right]\)
- C \([0,1]\)
- D \(\left[\dfrac{1}{3}, \dfrac{2}{9}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[-\dfrac{1}{3}, \dfrac{5}{9}\right]\)
Step-by-step Solution
Detailed explanation
For any event \(E\), the probability \(P(E)\) must satisfy \(0 \le P(E) \le 1\). For event \(A\), \(0 \le \dfrac{1}{3}(3x+1) \le 1\). Multiplying by \(3\), \(0 \le 3x+1 \le 3\). Subtracting \(1\), \(-1 \le 3x \le 2\). Dividing by \(3\), \(-\dfrac{1}{3} \le x \le \dfrac{2}{3}\).…
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