(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 $\quad$

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

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)

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