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Let $A$ and $B$ be two events, such that: $P(\overline{A \cup B} ) $$= \large\frac{1}{6}\:,\;\; P( A \cap B )$$= \large\frac{1}{4}$,$\;$and $\;\;P(\overline A)$$=\large\frac{1}{4} where \overline A stands for complement of event A. Then events A and B are (A) Independent but not equally likely \quad(B) Mutually exclusive and independent\quad (C) Equally likely and mutually exclusive \quad (D) Equally likely but not independent \quad Can you answer this question? 1 Answer 0 votes Toolbox: • P(A\cup B)=P(A)+P(B)-P(A\cap B) Given P( \overline{A \cup B}) = \large\frac{1}{6} P( A \cap B) = \large\frac{1}{4} P( \overline A) = \large\frac{1}{4} \Rightarrow P(A) = 1-P(\overline A) = \large\frac{3}{4} P( \overline{A \cup B}) = 1-P(A \cup B) \qquad\qquad = 1-P(A)-P(B)+P(A \cap B) \Rightarrow\: \large\frac{1}{6} = \large\frac{1}{4}$$-P(B)+\large\frac{1}{4}$
$\Rightarrow P(B)= \large\frac{1}{3}$
Since $P(A \cap B ) = P(A) \: P(B)\: and \: P(A) \neq P(B)$
$\therefore$ A and B are independent but not equally likely.
Ans : (A)
edited Mar 26, 2014