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

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  • $\begingroup$ 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. $\endgroup$ Apr 26, 2021 at 11:31

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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.

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

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

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