# How do I show that given $p(A \cap B) > 0$, when $p(A \cap B)$ decreases, and $p( B)$ remains the same, $p(A)$ must decrease?

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.

• 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\$ – David Diaz Mar 14 at 22:14
• 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. – dantopa Mar 14 at 22:16

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: $$A = B^c$$)