(In)dependence of pairs of events I'm currently working through my probability homework, and one of the questions is:
Two events are independent of each other when $P(A \cap B) = P(A) \cdot P(B)$. What can you say about the (in)dependence of the following pairs of events:
$$
\begin{align*}
&(a)\textrm{ } A \textrm{ and } \neg B\\
&(b)\textrm{ } \neg A \textrm{ and } B\\
&(c)\textrm{ } \neg A \textrm{ and } \neg B
\end{align*}
$$
What can you say about this? I'm not exactly sure what to do.
This is what I've managed now:
(a)
\begin{align*}
P(A \cap \neg B) &= P(A) - P(A \cap B)\\    
&= P(A) - P(A) \cdot P(B)\\
&= P(A) \cdot (1 - P(B))\\
&= P(A) \cdot P(\neg B) \textrm{, so they are independent}
\end{align*}
(b)
\begin{align*}
P(\neg A \cap B) &= P(B) - P(A \cap B)\\    
&= P(B) - P(A) \cdot P(B)\\
&= P(B) \cdot (1 - P(A))\\
&= P(\neg A) \cdot P(B) \textrm{, so they are independent}
\end{align*}
but I'm still stuck on (c).
 A: To give you a start.
If $A$ and $B$ are independent events then:$$P(A\cap B^c)=P(A)-P(A\cap B)=P(A)-P(A)P(B)=P(A)[1-P(B)]=P(A)P(B^c)$$
In the second equality the independence is used.
A: To expand on my comment to the question, if $A$ and $B$ are independent:
$$P(A^c\cap B^c)=P(A^c)-P(A^c\cap B)=1-P(A)-[P(B)-P(A\cap B)]\\=1-P(A)-P(B)+P(A)P(B)=(1-P(A))(1-P(B))=P(A^c)P(B^c)$$

However, you know that if $A$ and $B$ are independent, then $A^c$ and $B$ are independent, from the (a) you have proved.
Then, exchanging $A$ and $B$, that means that if $B$ and $A$ are independent, then $B^c$ and $A$ are independent. That's for the (b).
Then, replacing, in the statement of (a), $B$ by $B^c$, you have: if $A$ and $B^c$ are independent, then $A^c$ and $B^c$ are independent. But you already know that if $A$ and $B$ are independent, then $A$ and $B^c$ are independent (that's the (b)).
Therefore, if $A$ and $B$ are independent (then $A$ and $B^c$ are independent) then $A^c$ and $B^c$ are independent.

You have to understand that in a statement like "if $A$ and $B$ are independent, then $A^c$ and $B$ are independent", $A$ and $B$ are really "place holders", that you can replace with anything you want. You may as well read it as "if $U$ and $V$ are independent, then $U^c$ and $V$ are independent". The only constraint is that $U$ and $V$ are events, so that the sentence is meaningful.
A: The same approach you used in (a) would work, except with $A^c$ instead of $A$. Also, because you know $A^c$ and $B$ are independent:
$P(A^c \cap B^c) = $
$P(A^c) - P(A^c \cap B) = $
$P(A^c) - P(A^c)\times P(B) = P(A^c) (1 - P(B)) = P(A^c)P(B^c)$ 
