If A and B are two events such that <span class="math-tex">\(\rm P(A\cup B) = \dfrac{5}{6}\)</span>, <span class="math-tex">\(\rm P(A\cap B) = \dfrac{1}{3}\)</span> and <span class="math-tex">\(\rm P(\bar{B}) = \dfrac{1}{2}\)</span>, then the events A and B are:
\(P(B) = 1 - \frac{1}{2} = \frac{1}{2}\)
\(P(A) = P(A \cup B) - P(B) + P(A \cap B) = \frac{5}{6} - \frac{1}{2} + \frac{1}{3} = \frac{2}{3}\)
\(P(A)\,P(B) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} = P(A \cap B)\)
So A and B are independent. They are not mutually exclusive, since \(P(A \cap B) \ne 0\).
Answer: (B) Independent
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