I would like to know if the following proof implies that if A and B are independent, that does not mean that they are mutually exclusive:
If $A$ and $B$ are mutually exclusive then $A \cap B$ = $\emptyset$ $\Rightarrow$ P($A \cap B$)=0
We know that P($A|B$)=P($A \cap B$)/P($B$), and since $A$ and $B$ are independent, P($A$)=P($A \cap B$)/P($B$) and P($B$) =P($A \cap B$)/P($A$).
Since P($A \cap B$)=0, we immediately obtain that P($A$)=P($B$)=$\frac{0}{\:0}$. Seeing that that is impossible, we must derive that If $A$ and $B$ are independant, they must never be mutually exclusive.
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$\begingroup$Yes, that is basically it.
By definition: Independence of events $A, B$ means that $\mathsf P(A\cap B)=\mathsf P(A)\,\mathsf P(B)$
By definition: Mutual exclusion of events $A,B$ means that $A\cap B=\emptyset$ and thus implies that $\mathsf P(A\cap B)= 0$.
So if neither event has a probability measure of $0$, therefore they cannot be both independent and mutually exclusive.
Note: If one of the sets does have a probability measure of $0$, then they will be independent, and might be mutually exclusive.
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