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

First convert every given probability into a form that involves $$P(A)$$, $$P(B)$$ and $$P(A\cap B)$$.

1. Complement of the union: $$P(\overline{A\cup B})=\frac16 \;\Longrightarrow\; P(A\cup B)=1-\frac16=\frac56$$

2. Complement of $$A$$: $$P(\bar A)=\frac14 \;\Longrightarrow\; P(A)=1-\frac14=\frac34$$

3. Intersection is already given: $$P(A\cap B)=\frac14$$

Next use the addition law of probability: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

Substitute the known numbers:
$$\frac56 = \frac34 + P(B) - \frac14$$

Simplify the right‐hand side:
$$\frac34 - \frac14 = \frac12$$
So we get
$$\frac56 = \frac12 + P(B)$$

Isolate $$P(B)$$:
$$P(B)=\frac56-\frac12=\frac56-\frac36=\frac26=\frac13$$

Now check the required properties.

Equal likelihood: $$P(A)=\frac34,\;P(B)=\frac13$$ — they are not equal, so events are not equally likely.

Independence: events are independent if $$P(A\cap B)=P(A)\,P(B)$$.
Compute $$P(A)\,P(B)=\frac34\cdot\frac13=\frac14$$, which equals the given $$P(A\cap B)=\frac14$$. Hence $$A$$ and $$B$$ are independent.

Mutual exclusiveness: mutually exclusive events satisfy $$P(A\cap B)=0$$, but here $$P(A\cap B)=\frac14\ne0$$. So they are not mutually exclusive.

Therefore the correct description is: independent but not equally likely.

Option C which is: independent but not equally likely