# If $P(A|D) > P(A)$ and $P(B|D) > P(B)$, then is $P(A \cap B|D) > P(A \cap B)$?

There are 3 events $$A, B, D$$ such that $$D$$ makes $$A$$ more likely and $$D$$ makes $$B$$ more likely. Does this mean that $$D$$ makes it more likely that both $$A$$ and $$B$$ occur? How can you prove this using mathematical formalism to prove it if it’s true or what counter example makes it false?

• I helped you did formatting with MathJax this time. Going forward, please make sure your math expressions are formatted properly. Sep 18 at 15:21
• Draw a Venn diagram and attempt to construct a counterexample from what it shows you. Consider a case where $A\cap B \cap D=0.$
– whuber
Sep 18 at 15:27

If inequalities are strict, then counter examples are easy to make. For example, consider $$\Omega = \{1, 2, 3, 4, 5\}$$ where $$P(\{1\}) = P(\{2\}) = P(\{3\}) = P(\{4\}) = P(\{5\}) = 1/5$$. Let $$A = \{1, 2\}, B = \{4, 5\}$$, $$D = \{2, 4\}$$. Then $$P(A|D) = 1/2 > P(A) = 2/5$$, $$P(B|D) = 1/2 > P(B) = 2/5$$, while $$P(A \cap B|D) = 0 = P(A \cap B)$$, because $$A \cap B = \varnothing$$.
A more interesting question is thus whether $$P(A \cap B|D) \geq P(A \cap B)$$ holds given $$P(A|D) \geq P(A)$$ and $$P(B|D) \geq P(B)$$. This is still false, and we can tune the first counter example by making $$A \cap B \cap D = \varnothing$$ but pairwise intersection events $$A \cap D$$ and $$B \cap D$$ are not. This can be done by setting $$A = \{1, 3\}$$, $$B = \{3, 5\}$$, and $$D = \{1, 5\}$$, under which $$P(A|D) = 1/2 \geq P(A) = 2/5$$, $$P(B|D) = 1/2 \geq P(B) = 2/5$$, but $$P(A \cap B|D) = 0 < P(A \cap B) = 1/5$$.