Prove that if events A and B are independent, then the complement events of A and B are also independent. I know that:
$$\begin{gathered}P\left(A\cap B\right)=P\left(A\right)P\left(B\right)\\
P\left(A^{C}\right)=1-P\left(A\right)\\
P\left(B^{C}\right)=1-P\left(B\right)
\end{gathered}
$$
My proof so far:
$$\begin{gathered}P\left(A^{C}\cap B^{C}\right)=\left(1-P\left(A\right)\right)\left(1-P\left(B\right)\right)=\\
1-P\left(B\right)-P\left(A\right)+P\left(A\right)P\left(B\right)=1-P\left(B\right)-P\left(A\right)+P\left(A\cap B\right)
\end{gathered}
$$
After that, I'm stuck.  Any help would be appreciated.
 A: Assume $A$ and $B$ are independent. Then
\begin{align}
P(A^c \cap B^c) 
&= 1 - P(A \cup 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).
\end{align}
A: As you have found :
$$P(A') = 1-P(A)$$
$$P(B') = 1- P(B)$$
Now clearly $P(A')P(B') =1 - [ P(A) + P(B)]  + P(A \cap B)$
From set algebra we know that
$P(A) + P(B) = P(A \cup B) + P(A\cap B)$
Substituting, we have $P(A')P(B') = P([A\cup B]')$
Now from De morgans law we know that:
$[A\cup B]' = [A' \cap B']$
Substituting, we have $P(A')P(B') = P(A' \cap B')$ , as required.
A: gradient23's proof is great, in my opinion, but I would like to show another proof that seems more intuitive to me, though much less rigorous.
The proof is based on a verbal definition of independence from wikipedia:

two events are independent [...] if the occurrence of one does not affect the probability of occurrence of the other

In addition, we use the fact that independence is symmetric.
The (non-rigorous) proof:


*

*We assume that $A$ and $B$ are independent.

*By definition, the occurrence of  $A$ doesn't affect the probability of $B$.

*Thus, the occurrence of  $A$ also doesn't affect the probability of $B^C$.

*So by definition, $A$ and $B^C$ are also independent, which by definition again means that the occurrence of  $B^C$ doesn't affect the probability of $A$.
(Here we used the symmetry of independence.)

*Therefore, the occurrence of $B^C$ also doesn't affect the probability of $A^C$.

*So by definition, $B^C$ and $A^C$ are also independent.



One could convert the proof to the language of math:
$$\begin{gathered}A\text{ and }B\,\text{are independent}\\
\downarrow\\
P\left(B|A\right)=P\left(B\right)\\
\downarrow\\
1-P\left(B^{C}|A\right)=1-P\left(B^{C}\right)\\
\downarrow\\
P\left(B^{C}|A\right)=P\left(B^{C}\right)\\
\downarrow\\
A\text{ and }B^{C}\,\text{are independent}\\
\downarrow\\
P\left(A|B^{C}\right)=P\left(A\right)\\
\downarrow\\
1-P\left(A^{C}|B^{C}\right)=1-P\left(A^{C}\right)\\
\downarrow\\
P\left(A^{C}|B^{C}\right)=P\left(A^{C}\right)\\
\downarrow\\
B^{C}\text{ and }A^{C}\,\text{are independent}
\end{gathered}
$$
But now we used conditional probabilities, which might be a problem in case $P(A)=0$ or $P(B^C)=0$ (here is a discussion about this problem).
A: By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.
So, $P(A/B)=P(A \cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A \cap B)= P(A)P(B)$.
Hence proved.
