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I can't seem to wrap my head around why $P(a \text{ and } b)$ is minimized when $P(a \text{ or } b)$ is maximized. This comes from PIE:

$$P(a \text{ or } b) = P(A) + P(B) - P(a \text{ and } b).$$

Can someone please explain the intuition behind this? I'm even trying to picture the Venn diagram in my head, but this exact relationship doesn't make sense.

Xander Henderson
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Eric Aldrin
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  • If you fix $P(A),P(B)$ then the way to maximize $P(A\cup B)$ is to make sure you remove the least possible when substracting $P(A\cap B)$. That is, minimizing $P(A\cap B)$. – LeviathanTheEsper Jun 18 '20 at 22:06
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    Your title does not match your question, by the way (although fundamentally it makes no difference). – Andrew Chin Jun 18 '20 at 22:08
  • It's because of the negative sign of P(A and B). When P(A or B) is maximized, we have that P(A) + P(B) is maximized (because it's a positive quantity), while P(A and B) is minimized. – user821 Jun 18 '20 at 22:09

2 Answers2

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$A \cap B$ is the overlapping part in $A \cup B$.

Imagine you have two sheets of paper and you want to maximize their surface.

If they overlap, you can increase the surface by making them overlap less.

Philipp123
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In the relation $$x=y-z,$$ you get the largest value for $x$ when the value of $z$ is a minimum. That is, you would get the largest difference between two numbers if the number you are subtracting is made as small as possible.

Andrew Chin
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