Let A and B be two events such that P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} and P(\overline{A})=\frac{1}{4}. Then events A and B are

Correct Answer is : independent but not equally likely.

mutually exclusive and independent.

equally likely and mutually exclusive.

equally likely but not independent.

NIMCET 2009 Question Paper with solutions

Question : 17 of 120

Marks: +1, -0

Let

A

and

B

be two events such that

P(\ov{A ∪ B})=\;{1}/{6}, P(A ∩ B)=\;{1}/{4}

and

P(\ov{A})=\;{1}/{4}

. Then events A and B are

Solution:

P(\ov{A ∪ B})=\;{1}/{6}

⇒ \;\; H(A ∪ B)=1-\;{1}/{6}=\;{5}/{6}

P(\ov{A})=\;{1}/{4} ⇒ P(A)=1-\;{1}/{4}=\;{3}/{4}

P(B)=P(A ∪ E)+P(A ∩ B)-P(A)

P(B)=P(A ∪ E)+P(A ∩ B)-P(A)

=\;{5}/{6}+\;{1}/{4}-\;{3}/{4}=\;{1}/{3}

⇒ \;\; P(A)=\;{3}/{4}: P(B)=\;{1}/{3}

∵ P(A ∩ B)=P(A) ⋅ P(B)

\;{1}/{4}=\;{3}/{4} ⋅ \;{1}/{3}=\;{1}/{4}

∴ A

and

B

are independent but not equally likely

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