Let $$A$$ and $$B$$ be two events such that $$P\left(\overline{A \cup B}\right) = \frac{1}{6}$$, $$P(A \cap B) = \frac{1}{4}$$ 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:

We are given three probability values involving the events $$A$$ and $$B$$:

$$P\!\left(\overline{A \cup B}\right)=\dfrac16, \qquad P(A\cap B)=\dfrac14, \qquad P(\bar A)=\dfrac14.$$

First, we recall the complement rule:

$$P(\overline{E}) = 1 - P(E) \quad\text{for any event }E.$$ Applying this to the event $$A\cup B,$$ we write

$$P(A\cup B)=1-P\!\left(\overline{A\cup B}\right)=1-\dfrac16=\dfrac56.$$ So, $$P(A\cup B)=\dfrac56.$$

Next, we use the complement of $$A$$ to obtain $$P(A).$$ By the same rule,

$$P(A)=1-P(\bar A)=1-\dfrac14=\dfrac34.$$

To find $$P(B),$$ we employ the addition (union) formula for two events:

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

Substituting the values already known, we have

$$\dfrac56=P(A)+P(B)-P(A\cap B).$$

Now we insert $$P(A)=\dfrac34$$ and $$P(A\cap B)=\dfrac14$$:

$$\dfrac56=\dfrac34+P(B)-\dfrac14.$$

We simplify the right-hand side step by step. First combine the two fractions involving quarters:

$$\dfrac34-\dfrac14=\dfrac{3-1}{4}=\dfrac24=\dfrac12.$$

So the equation becomes

$$\dfrac56=\dfrac12+P(B).$$

To isolate $$P(B),$$ subtract $$\dfrac12$$ from both sides:

$$P(B)=\dfrac56-\dfrac12.$$

We bring both fractions to a common denominator of $$6$$:

$$\dfrac56-\dfrac12=\dfrac56-\dfrac36=\dfrac{5-3}{6}=\dfrac26=\dfrac13.$$

Thus, $$P(B)=\dfrac13.$$

With $$P(A)=\dfrac34$$ and $$P(B)=\dfrac13,$$ we can now investigate the required properties of the events.

(i) Testing for independence. Two events $$A$$ and $$B$$ are independent when the following condition holds:

$$P(A\cap B)=P(A)\,P(B).$$

We calculate the product on the right:

$$P(A)\,P(B)=\left(\dfrac34\right)\!\left(\dfrac13\right)=\dfrac{3}{4}\cdot\dfrac{1}{3}=\dfrac14.$$

The left side, given in the question, is $$P(A\cap B)=\dfrac14.$$ Both sides are equal:

$$P(A\cap B)=P(A)\,P(B)=\dfrac14.$$

Therefore, the events $$A$$ and $$B$$ are independent.

(ii) Checking whether the events are equally likely. Events are called equally likely when their probabilities are equal, that is, when $$P(A)=P(B).$$ Here, $$P(A)=\dfrac34$$ and $$P(B)=\dfrac13,$$ which are clearly different. Hence $$A$$ and $$B$$ are not equally likely.

(iii) Checking whether the events are mutually exclusive. Events are mutually exclusive if they cannot occur together, which translates to

$$P(A\cap B)=0.$$

But we know $$P(A\cap B)=\dfrac14\neq0,$$ so $$A$$ and $$B$$ are not mutually exclusive.

Combining all the above conclusions, the events are independent but not equally likely.

Hence, the correct answer is Option A.