Events $A$ and $B$ are such that $P(A)=\frac{1}{2}$,$P(B)=\frac{7}{12}$ and $P(\text{not } A \text{ or not } B)=\frac{1}{4}$. State whether $A$ and $B$ are independent?

(B) It is given that $P(A)=\frac{1}{2}$,$P(B)=\frac{7}{12}$ and $P(A' \cup B')=\frac{1}{4}$.
Using De Morgan's Law,$A' \cup B' = (A \cap B)'$.
So,$P((A \cap B)') = \frac{1}{4}$.
Since $P(E') = 1 - P(E)$,we have $1 - P(A \cap B) = \frac{1}{4}$.
Therefore,$P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$.
Now,calculate $P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$.
Since $P(A \cap B) = \frac{3}{4}$ and $P(A) \times P(B) = \frac{7}{24}$,we observe that $P(A \cap B) \neq P(A) \times P(B)$.
Thus,events $A$ and $B$ are not independent.