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Just asking for a formal proof.

Assume $p(A \cap B) > 0$. And assume that there can be four possibilities: $(A \cap B)$, $(A^{c} \cap B)$, $(A \cap B^{c})$, $(A^{c} \cap B^{c})$. Let’s say $p(A \cap B)$ decreases. But $p(B)$ stays the same. How to I prove that $p(A)$ must decrease? Or is this wrong?

I’m thinking of using $p(A \cap B) = p(A) + p(B) - p(A \cup B)$. But I don’t know how to proceed.

Thanks.

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  • $\begingroup$ Seems like a good start. Are $a$ and $b$ independent? If not, this isnt necessarily true. For texxing variables, logical symbols, etc. try enclosing the following symbols inside of dollar signs like so: \$P(a \cap b)\$, \$P(a \cup b)\$, \$\lnot a \land b\$, \$a \land \sim b\$ $\endgroup$ – David Diaz Mar 14 at 22:14
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. $\endgroup$ – dantopa Mar 14 at 22:16
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This answer is based on the comment by @DavidDiaz

The claim is true if $A,B$ are independent (both before and after the "decrease"), because then $P(A\cap B) = P(A)P(B)$. So holding $P(B)$ constant, $P(A)$ and $P(A \cap B)$ must decrease together, in fact by the same proportion.

The claim can be false if $A,B$ are dependent. Here is a counterexample:

  • Before: $P(A) = P(B) = 0.3, P(A \cap B) = 0.09$ (they were independent)

  • After: $P(B) = 0.3, P(A) = 0.7, P(A \cap B) = 0$ (now they become complementary: $A = B^c$)

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