Suppose sets $A , B,$ and $D$ are independent. Is it guaranteed that $A \cap B^c \cap D$ is independent from $B^c \cup D^c$?
Isn't $B^c$ ($B$ complement, or $B$ not happening) giving me information about $B^c \cup D^c$? (A) I know that $B$ complement happened so it should increase the chances of $B^c \cup D^c$ happening, thus they are not independent.
Is my statement (A) correct?
For inspiration, let's examine Venn diagrams of each set.
Elements of $\mathcal{X}=A\cap B^c\cap D$ are (a) in $A;$ (b) not in $B;$ and (c) in $D$. This region is highlighted in yellow.
Elements of $\mathcal{Y}=B^c\cup D^c$ are either not in $B$ or not in $D$ (or both). This includes a lot of the diagram, so instead I have highlighted its complement -- namely, everything that is not in this set. The complement of $\mathcal Y$ is $B\cap D$ (this evident equality is one of DeMorgan's Laws):
Now, two sets $\mathcal{X}$ and $\mathcal Y$ are independent if and only if $\mathcal{X}$ and $\mathcal{Y}^c$ are independent. But since $\mathcal{X}$ and $\mathcal{Y}^c$ are disjoint, the probability of their intersection is zero. When both $\mathcal X$ and $\mathcal{Y}^c$ have nonzero probability, the rule for independence won't hold.
From this insight we may construct an example where the two sets $\mathcal X$ and $\mathcal Y$ are not independent. (Once we have found this example, we needn't repeat any of the foregoing analysis: the example alone is enough to resolve the problem. But I thought you might want to understand how such examples can be cooked up and what kind of thinking goes into that.)
For this example I will put one element into $\mathcal{X}$ and one into $\mathcal{Y}^c:$ since neither $\mathcal X$ nor $\mathcal{Y}^c$ can be empty, yet are disjoint, that's as simple as an example possibly could be.
Looking at the two diagrams, then, let's put (say) the number $0$ into $A$ and $D$ but not $B$ and (say) the number $1$ into both $B$ and $D.$ One way to do this is to define
$$A = \{0\},\quad B = D = \{1\}.$$
Define a probability measure for $A\cup B \cup D = \{0,1\}$ by giving $\{0\}$ and $\{1\}$ equal chances of $1/2.$ Note that any set in this probability space can have probability $0,$ $1/2,$ or $1,$ but no other value is possible. Here's a Venn diagram showing the set elements:
Computing
$$\Pr(\mathcal{X}) = \Pr(A\cap B^c\cap D) = \Pr(\{0\}) = \frac{1}{2},$$
$$\Pr(\mathcal{Y}) = \Pr(B^c\cup D^c) = \Pr(\{0\}) = \frac{1}{2},$$
we see that the independence of $\mathcal{X}$ and $\mathcal{Y}$ would mean (by definition)
$$\Pr(\mathcal{X}\cap\mathcal{Y}) = \Pr(\mathcal{X})\Pr(\mathcal{Y}) = \frac{1}{2}\times \frac{1}{2} = \frac{1}{4}.$$
Since a probability of $1/4$ is impossible, $\mathcal{X}$ and $\mathcal{Y}$ cannot be independent.
