# Why is $P(a \text{ and } b)$ maximized when $P(a \text{ or } b)$ is minimized?

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.

• 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 $P(A\cap B)$. Commented Jun 18, 2020 at 22:06
• Your title does not match your question, by the way (although fundamentally it makes no difference). Commented Jun 18, 2020 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. Commented Jun 18, 2020 at 22:09

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