# 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 $$A^c \cap B \Longleftrightarrow x \notin A \land x \in B$$.

• It is neither true that $A^{C}\cap B=A \setminus (A \cap B)$ nor is it true that these two events have the same probability. Apr 26, 2021 at 11:31

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.

$$(\Omega\setminus E_1) \cap E_2\cap...\cap E_k= E_2\cap...\cap E_k \setminus E_1\cap E_2\cap...\cap E_k$$. 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.

• Where does the first equality come from? Apr 26, 2021 at 11:39
• @EpsilonAway Verify that each side is contained in the other. It is quite simple. Apr 26, 2021 at 11:40