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 the complement of the event A. Then the events A and B are

(C) independent but not equally likely

(A) mutually exclusive and independent

(B) equally likely but not independent

(D) independent and equally likely

JEE Main · 2014 · 06 Apr 2014 · 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 the complement of the event A. Then the events A and B are

Answer: (C)

From

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

we get

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

, and from

P (\overset{ˉ}{A}) = \frac{1}{4}

we get

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

. Then

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

gives

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

, so

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

. Testing independence,

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.

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option