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

Correct option is 4independent but not equally likely

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

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Correct option is D

We have,P(A∪B¯)=16,P(A∩B)=14P(A¯)=14⇒P(A)=1−14=34∴ P(A∪B→)=1−P(A∪B)=1−[P(A)+P(B)−P(A∩B)]⇒ 16=1−34−P(B)+14⇒ P(B)=1−12−16=6−3−16=26=13⇒PA∩B=PAPB=14

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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

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Ready to Test Your Skills?

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