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

Question

Let $ A $ and $ B $ are two events such that $ P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} $ and $ P(\bar{A})=\frac{1}{4} $, where $ \bar{A} $ represents the complement of event $ A $. Then, events $ A $ and $ B $ are, mutually exclusive and independent., independent but not equally likely., equally likely but not independent., equally likely and mutually exclusive.

MARKS App · Accepted Answer

Given, P A ∪ B ¯ = 1 6 ⇒ 1 - P A ∪ B = 1 6 ⇒ 1 - P A - P B + P A ∩ B = 1 6 ⇒ P ( A ¯ ) - P B + P A ∩ B = 1 6 ⇒ 1 4 - P B + 1 4 = 1 6 ⇒ P B = 1 3 and P A = 1 - 1 4 = 3 4 As, P A ∩ B = P A P B So, events A and B are independent events but they are not equally likely.

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