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 \(P(\bar{A})=\frac{1}{4}\), where \(\bar{A}\) stands for complement of event \(A\). Then events \(A\) and \(B\) are

(C) independent but not equally likely

(A) equally likely and mutually exclusive

(B) equally likely but not independent

(D) mutually exclusive and independent

JEE Main · 2005 · 08 May 2005 · Offline · Probability

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 (\overset{ˉ}{A}) = \frac{1}{4}

, where

\overset{ˉ}{A}

stands for complement of event

A

. Then events

A

and

B

are

Answer: (C)

From

P (\overset{ˉ}{A}) = \frac{1}{4}

we get

P (A) = \frac{3}{4}

, and from

P (\overline{A \cup B}) = \frac{1}{6}

we get

P (A \cup B) = \frac{5}{6}

. Using

P (A \cup B) = P (A) + P (B) - P (A \cap B)

with

P (A \cap B) = \frac{1}{4}

, we have

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

, so

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

. Now

P (A) P (B) = \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4} = P (A \cap B)

, so the events are independent; but

P (A) = \frac{3}{4} \neq = \frac{1}{3} = P (B)

, so they are not equally likely. The events are independent but not equally likely.

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option