1. Given data

We have: $P\big(\overline{A\cup B}\big)=\frac16,$ $P(A\cap B)=\frac14,$ $P(\overline A)=\frac14.$

We need to determine the relation between events $A$ and $B$.

2. Find $P(A)$

Since $P(\overline A)=\frac14,$ we get $P(A)=1-\frac14=\frac34.$

3. Find $P(A\cup B)$

Using complement: $P(A\cup B)=1-P\big(\overline{A\cup B}\big)=1-\frac16=\frac56.$

4. Use the addition formula to find $P(B)$

We know $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ So, $\frac56=\frac34+P(B)-\frac14.$ Simplify: $\frac56=\frac12+P(B).$ Hence, $P(B)=\frac56-\frac12=\frac{5-3}{6}=\frac26=\frac13.$

Thus, $P(B)=\frac13.$

5. Check whether $A$ and $B$ are equally likely

Two events are equally likely if $P(A)=P(B).$ But here, $P(A)=\frac34,\qquad P(B)=\frac13.$ Since $\frac34\ne\frac13,$ $A$ and $B$ are not equally likely.

6. Check whether they are mutually exclusive

Mutually exclusive means $P(A\cap B)=0.$ But given, $P(A\cap B)=\frac14\ne 0.$ So they are not mutually exclusive.

7. Check whether they are independent

Independent events satisfy $P(A\cap B)=P(A)P(B).$ Now, $P(A)P(B)=\frac34\cdot\frac13=\frac14.$ And given, $P(A\cap B)=\frac14.$ So, $P(A\cap B)=P(A)P(B),$ therefore $A$ and $B$ are independent.

8. Conclusion

• Not equally likely
• Independent
• Not mutually exclusive

Hence the correct option is: $\boxed{\text{C: independent but not equally likely}}$