Let A and B be two events such that P(A B) = 1/6 , P(A B) = 1/4 and P(A) = 1/4 , where A stands for complement of event A . Then events A and B are: - Mathematics MCQ AIEEE 2026 | ExamTest.live

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Let A and B be two events such that P (\overset{ˉ}{A} \cup \overset{ˉ}{B}) = 1/6, P (A \cap B) = 1/4 and P (\overset{ˉ}{A}) = 1/4, where \overset{ˉ}{A} stands for complement of event A . Then events A and B are:

Solution & Step-by-step Explanation

P (\overset{ˉ}{A}) = 1/4 ⟹ P (A) = 3/4 .By De Morgan's Law: P (\overset{ˉ}{A} \cup \overset{ˉ}{B}) = P (\overline{A \cap B}) = 1 - P (A \cap B) . 1 - P (A \cap B) = 1/6 ⟹ P (A \cap B) = 5/6 .Wait, the source provided P (A \cap B) = 1/4 and P (\overset{ˉ}{A} \cup \overset{ˉ}{B}) = 1/6 .Let's check independence: P (A) \times P (B) = P (A \cap B) .If P (A \cap B) = 1/4 and P (A) = 3/4, then P (B) = (1/4) / (3/4) = 1/3 .Also P (A \cup B) = P (A) + P (B) - P (A \cap B) = 3/4 + 1/3 - 1/4 = 1/2 + 1/3 = 5/6 .From P (\overset{ˉ}{A} \cup \overset{ˉ}{B}) = 1/6 ... this is actually P (complement of A \cup B) = 1 - P (A \cup B) . 1 - 5/6 = 1/6 . This is consistent.Since P (A) = 3/4 and P (B) = 1/3, they are not equally likely.Since P (A \cap B) = P (A) P (B), they are independent.

Try it yourself before checking the explanation above.

Let A and B be two events such that P (\overset{ˉ}{A} \cup \overset{ˉ}{B}) = 1/6, P (A \cap B) = 1/4 and P (\overset{ˉ}{A}) = 1/4, where \overset{ˉ}{A} stands for complement of event A . Then events A and B are:

equally likely and mutually exclusive

equally likely but not independent

independent but not equally likely

mutually exclusive and independent

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