# Conditional probability: theoretical exercise

I am trying to show that

$P(E\mid E\bigcup F) \geq P(E \mid F)$.

This is intuitively clear. But when expanding I get $P(E)\ P(F)\geq P(E\bigcup F)\ P(E \bigcap F)$. How to continue?

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Let $a=P(E\cap F^c)$, $b=P(E\cap F)$ and $c=P(F\cap E^c)$. You have that $P(E)=a+b$, $P(F)=b+c$. Since $E\cup F=((E\cap F^c)\cup(E\cap F)\cup (F\cap E^c))$ and since the the union is disjoint you have that $P(E\cup F)=a+b+c$. Therefore, the problem you stated reduces to showing $(a+b)(b+c)\geq b(a+b+c)$ which follows trivially since $ac=P(E\cap F^c)P(F\cap E^c)\geq 0$.
Hint: the difference between these two terms is $P(E\cap F^c)P(F\cap E^c)$.