COMEDK · Maths · 36. Probability
Let \(\mathrm{A}\) and \({B}\) be two events such that \(P(A / B)=\dfrac{1}{2}\) and \(P(B / A)=\dfrac{1}{3}\) and \(P(A \cap B)=\dfrac{1}{6}\) then, which one of the following is not true?
- A \(P\left(A^{\prime} \cap B\right)=\dfrac{1}{6}\)
- B \(A \text { and } B \text { are independent }\)
- C \(\mathrm{A} \text { and } \mathrm{B} \text { are not independent }\)
- D \(P(A \cup B)=\dfrac{2}{3}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{A} \text { and } \mathrm{B} \text { are not independent }\)
Step-by-step Solution
Detailed explanation
Given \(P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1}{2}\) and \(P(A \cap B) = \dfrac{1}{6}\), we have \(\dfrac{1/6}{P(B)} = \dfrac{1}{2}\), which implies \(P(B) = \dfrac{1}{3}\). Given \(P(B|A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{1}{3}\) and \(P(A \cap B) = \dfrac{1}{6}\), we…
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