1. Given probabilities

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

We first find $P(A)$ and $P(A\cup B)$.

2. Find $P(A)$

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

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

Since $P\big(\overline{A\cup B}\big)=1-P(A\cup B),$ we get $P(A\cup B)=1-\frac16=\frac56.$

4. Use union formula to find $P(B)$

We know $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Substitute the known values: $\frac56=\frac34+P(B)-\frac14.$ Now, $\frac34-\frac14=\frac12,$ so $\frac56=\frac12+P(B).$ Hence, $P(B)=\frac56-\frac12=\frac{5-3}{6}=\frac26=\frac13.$

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

5. Check if $A$ and $B$ are independent

For independence, we need $P(A\cap B)=P(A)P(B).$ Now, $P(A)P(B)=\frac34\cdot\frac13=\frac14.$ But $P(A\cap B)=\frac14.$ So, $P(A\cap B)=P(A)P(B).$ Hence, $A$ and $B$ are independent.

6. Check if $A$ and $B$ are equally likely

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

7. Check options

• A: independent but not equally likely. ✅

• B: independent and equally likely. ❌

• C: mutually exclusive and independent. ❌

  Mutually exclusive would require $P(A\cap B)=0,$ but here $P(A\cap B)=\frac14\ne 0.$

• D: equally likely but not independent. ❌

Therefore, the correct option is: $\boxed{\text{A}}$