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If $A$ and $B$ aren't disjoint and $A \cup B \neq \Omega$, then is $P(A \cup B) \geq P(A)P(B)$?

My only idea is to use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ but there's a minus in front of the intersection and the events don't have to be independent.

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    $\begingroup$ This gives the pretty result $P(A)P(B) \leq P(A \cup B) \leq P(A)+P(B)$. It is true in general, with the first being an equality when $P(A)=P(B)=1$ or $P(A)=P(B)=0$, and the second when $P(A \cap B)=0$. $\endgroup$
    – Henry
    Commented Apr 1 at 11:06

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The short answer is yes, and I believe that the reason is more fundamental than you suggest.

Since $P(A\cup B)\geq P(A)$ and $0 \leq P(B) \leq 1$ it follows quite naturally that $P(A) \geq P(A)\times P(B)$ and hence $P(A\cup B)\geq P(A)P(B)$

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