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

or, $$1 - P(A \cup B) = {1 \over 6}$$

$$\therefore$$ $$P(A \cup B) = 1 - {1 \over 6} = {5 \over 6};$$ $$P(A \cap B) = {1 \over 4}$$; and $$P(\overline A ) = {1 \over 4};$$

$$\therefore$$ $$P(A) = 1 - P(\overline A ) = 1 - {1 \over 4} = {3 \over 4}$$

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

or, $${5 \over 6} = {3 \over 4} + P(B) - {1 \over 4}$$

or, $$P(B) = {5 \over 6} - {1 \over 2} = {1 \over 3}$$

Now, $$P(A)\,.\,P(B) = {3 \over 4}.\,{1 \over 3} = {1 \over 4} = P(A \cap B)$$

i.e., events A and B are mutually independent.

Since the probability of A and B are different, so they are not equally likely events.

Therefore, (A) is the correct option.