$P\left(\overline{A\cup B}\right)=\frac{1}{6}$         

$\therefore P\left(A\cup B\right)=\frac{5}{6}$

$P\left(A\cap B\right)=\frac{1}{4}$

$P\left(\overline{A}\right)=\frac{1}{4}$         

$\therefore P\left(A\right)=\frac{3}{4}$

$\therefore P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$\therefore \frac{5}{6}=\frac{3}{4}+P\left(B\right)-\frac{1}{4}$

$\therefore P\left(B\right)=\frac{1}{3}$

$\therefore P\left(A\right)·P\left(B\right)=\frac{1}{3}\times \frac{3}{4}=\frac{1}{4}=P\left(A\cap B\right)$

Therefore, $A$ and $B$ are independent and since $P\left(A\right)\ne P\left(B\right)$ they are not equally likely.