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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.

johnsmith
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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 $P(A\cup B)\geq P(A)P(B)$

Red Five
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    Wow! How simple! I had reduced it to $1-P(A \cap B) \ge (1-P(A))(1-P(B)) $ and was trying to work with that. – marty cohen Mar 31 '24 at 03:55
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    The negative is an issue with any attempt like that. When I first read the question my initial reaction was to consider it a numerical rather than algebraic problem and see where that went. Luckily, the answer dropped out pretty quickly. Which makes it a rather nice question and I'm going to use it myself. So thanks. – Red Five Mar 31 '24 at 04:03