How to show that $\mathrm{Pr}(A^c \cap B) = \mathrm{Pr}(A \setminus (A\cap B))$ for independent events? I am trying to show that if $E_1,\dots,E_k$ are independent events, then so are $E_1^c,E_2,\dots,E_k$. Currently I am stuck at writing down what $\mathrm{Pr}((\Omega \setminus E_1)\cap E_2 \cap \dots E_k)$ evaluates to, where $\Omega$ is the universe. One source claimed that for sets $A$ and $B$, $A^c \cap B = A \setminus (A \cap B)$, but I have not been able to show this either, since I don't see how $A \setminus (A \cap B)$ follows from $A^c \cap B \Longleftrightarrow x \notin A \land x \in B$.
 A: First of all, $A^c \cap B = A \setminus (A \cap B)$ is not true. Correct statement such as this is $A^c \cap B = B \setminus (A \cap B)=B\setminus A$ (it is simple when you draw the  Venn diagrams, also $A^c \cap B$ means that element is in $B$ but not in $A$).
Anyway, you want to show that if $P(A\cap B)=P(A)P(B)$ then also  $P(A^c\cap B)=P(A^c)P(B)$, which is equal to $(1-P(A))P(B)$. Yes, you do have $n$ events, but here you can simply take $B=E_2\cap\dots\cap E_n$ and $A=E_1$, so considering only two events is enough.
How to do that? Measures in general fulfill the additive property, i.e. for disjoint events $C,D$ holds $P(C\cup D)=P(C)+P(D)$. You can use it here such as this:
$$
P(B)=P((B\cap A) \cup (B\setminus A) ) = P(B\cap A)+P(B\setminus A).
$$
Finally, only adding everything together, we get
$$
P(A^c\cap B)=P(B\setminus A) = P(B)-P(B\cap A) = P(B) - P(B)P(A)=(1-P(A))P(B),
$$
what you wanted to prove.
A: $(\Omega\setminus E_1) \cap E_2\cap...\cap E_k= E_2\cap...\cap E_k \setminus E_1\cap E_2\cap...\cap E_k$. So, $Pr (\Omega\setminus E_1) \cap E_2\cap...\cap E_k))=Pr(E_2\cap...\cap E_k)-Pr(E_1\cap E_2\cap...\cap E_k)$. Now use independence to finish the proof.
