If A and B are independent events such that 0 < P(A) < 1 and 0 < P(B)…

If A and B are independent then $P(A∩B)=P(A)P(B)$

now $P(A∩\overline{B})=P(A)−P(A∩B)=P(A)−P(A)P(B)$

$=P(A)\begin{Bmatrix}1−P(B)\end{Bmatrix}=P(A)P(\overline{B})$

Hence A and $\overline{B}$ are independent.

Similarly $P(\overline{A}∩\overline{B})=1−P(A∪B)=1−P(A)−P(B)+P(A∩B)$

$=1−P(A)−P(B)+P(A)P(B)=1−P(A)−P(B)\begin{Bmatrix}1−P(A)\end{Bmatrix}$

$=\begin{Bmatrix}1−P(A)\end{Bmatrix}\begin{Bmatrix}1−P(B)\end{Bmatrix}=P(\overline{A})P(\overline{B})$

Hence $\overline{A}$ and $\overline{B}$ are independent.

$P(\frac{A}{B})+P(\frac{\overline{A}}{B})=\frac{P(A∩B)}{P(B)}​+\frac{P(\overline{A}∩B)}{P(B)}​=P(A)+P(\overline{A})=1$

So, $P(A∩B) = P(A)P(B) = 0$  since $A/Q P(A) > 0$ and $P(B)>0$

A and B are not mutually exclusive.

So, A and B' are independent, A' and B are independent, A' and B' are independent

So option D is correct.