Proving basic probability inequalities So the title is a bit vague, but I am trying to solve some basic probability inequality. Given that $P(B\cap C) > P(B)P(C)$, how can I show that $P(B|C) > P(B|W\setminus C)$?
I started by showing that $P(B|C) > P(B)$, but I am not sure if this helps. Thoughts?
Thanks.
 A: $W$ must be the sample space, so that $W\setminus C = C^\complement$.
Indeed, stating $\mathbb P(B\cap C) >\mathbb P(B)\,\mathbb P(C)$ implies that $0<\mathbb P(C)<1$*, thus, from the definition of conditional probability,
$$\begin{align}
\mathbb P(B\,|\,C)\,\mathbb P(C) &> \mathbb P(B)\,\mathbb P(C)\\
\mathbb P(B|C) &> \mathbb P(B)\quad.
\end{align}$$
Also, adding the probability of $B\cap C^\complement$ to both sides of the inequality given, using that $\mathbb P(C^\complement)\neq0$,
$$\begin{align}
\mathbb P(B\cap C) + \mathbb P(B\cap C^\complement) &>\mathbb P(B)\,\mathbb P(C) + \mathbb P(B\cap C^\complement)\\
\mathbb P(B) &>\mathbb P(B)\,\mathbb P(C) + \mathbb P(B\cap C^\complement)\\
\mathbb P(B)[1 - \mathbb P(C)] &> \mathbb P(B\cap C^\complement)\\
\mathbb P(B)\,\mathbb P(C^\complement) &> \mathbb P(B\cap C^\complement)\\
\mathbb P(B) &> \mathbb P(B\,|\,C^\complement)\quad,
\end{align}$$
and we combine the two inequalities obtained to prove that $\mathbb P(B\,|\,C) > \mathbb P(B\,|\,C^\complement)$.
* If $\mathbb P(C)=0$, then $\mathbb P(B\cap C) > 0$, but since $B\cap C\subset C$, $\mathbb P(B\cap C)\le\mathbb P(C)$, so $\mathbb P(B\cap C) = 0$, a contradiction. On the other hand, if $\mathbb P(C)=1$, we would have $\mathbb P(B\cap C) > \mathbb P(C)$, which contradicts $\mathbb P(B\cap C)\le\mathbb P(C)$ as well.
A: By hypothesis, $P(B\cap C)>P(B)P(C)$, which is equivalent to $P(B\mid C)>P(B)$. The Law of Total Probability tells us that $P(B)=P(B\mid C)P(C)+P(B\mid C^c)P(C^c)<P(B\mid C)$, by hypothesis, yielding $P(B\mid C)>P(B\mid C^c)$.
