Let A and B two events such that $\mathrm{P}\left(\overline{\mathrm{A}\cup \mathrm{B}}\right)=\frac{1}{6}, \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{1}{4}$ and $\mathrm{P}\left(\overline{\mathrm{A}}\right)=\frac{1}{4}$ where $\overline{\mathrm{A}}$ stands for the complement of the event A then the events A and B are

$\begin{array}{l}\mathrm{P}(\mathrm{A}\cup \mathrm{B})=1-\frac{1}{6}=\frac{5}{6},\mathrm{P}\left(\mathrm{A}\right)=1-\frac{1}{4}=\frac{3}{4}\\ P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\\ \frac{5}{6}=\frac{3}{4}+\mathrm{P}\left(\mathrm{B}\right)-\frac{1}{4}\\ \mathrm{P}\left(\mathrm{B}\right)=\frac{1}{3}\end{array}$
$\text{ Now }\mathrm{P}\left(\mathrm{A}\right)\cdot \mathrm{P}\left(\mathrm{B}\right)=\frac{3}{4}\cdot \frac{1}{3}=\frac{1}{4}=\mathrm{P}(\mathrm{A}\cap \mathrm{B})$
$\therefore \mathrm{A},\mathrm{B}\text{ are independent but not equally likely. }$