Let A and B be two events such that , P \left(\right. \bar{A \cup B} \left.\right) \textrm{ }\textrm{ } = \textrm{ }\textrm{ } \frac{1}{6} , P \left(\right. A \cap B \left.\right) \textrm{ }\textrm{ } = \textrm{ }\textrm{ } \frac{1}{4} and , P \left(\right. \bar{A} \left.\right) \textrm{ }\textrm{ } = \textrm{ }\textrm{ } \frac{1}{4} , \textrm{ }\textrm{ } where \textrm{ }\textrm{ } \bar{A} \textrm{ }stands for the complement of the event A. Then the events A and B are

Engineering

Mathematics

Addition Theorem on Probability

Question

Let A and B be two events such that , $P (\bar{A \cup B}) = \frac{1}{6}, P (A \cap B) = \frac{1}{4}$ and , $P (\bar{A}) = \frac{1}{4}, where \bar{A}$ stands for the complement of the event A. Then the events A and B are

mutually exclusive and independent

independent and equally likely

independent but not equally likely

equally likely but not independent

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$P (A \cup B) = \frac{5}{6}; P (A) = \frac{3}{4}$

P(A $\cup$ B) = P(A) + P(B) – P(A $\cap$ B)

$\frac{5}{6} = \frac{3}{4} + P (B) - \frac{1}{4}$

$P (B) = \frac{1}{3}$

Now, P(A $\cap$ B) = P(A).P(B)

Hence, A and B are independent events