How to prove if $P(B|A) > P(B)$ then $P(B|A') < P(B)$ I hate proof questions, I never know how to start. My question says:

Let $A$ and $B$ be two events in a sample space such that $0 < P(A) < 1$. Let $A'$ denote the complement event of $A$. Show that if $P(B|A) > P(B)$ then $P(B|A') < P(B)$.

How do I go about answering this question? I don't understand why $P(B)$ is between $1$ and $0$ but not $P(B)$?
Has it got something to do with using that Bayesian Theorem?
 A: No need for Bayes. 
Observe that by the definition of conditional probability:
$$ P(B|A)P(A) = P(A \text{ and } B) $$
and that 
$$ P(A \text{ and }B) + P(A' \text{ and } B) = P((A \text{ or }A') \text{ and } B) = P(B) $$
So we have
$$ P(B) = P(B|A) P(A) + P(B|A') P(A') \tag{%}$$
If $P(B|A)> P(B)$ and $P(B|A') > P(B)$, equation (%) implies that
$$ P(B) > P(B) P(A) + P(B) P(A') = P(B)\left[ P(A) + P(A')\right] = P(B) $$ 
which is a contradiction. 

Intuitively this is the statement that "the weighted average of two numbers must lie between the two numbers". $P(B)$, the probability of $B$ happening over the whole sample space, is the weighted average of $P(B|A)$, the probability of $B$ happening over the sample space for which $A$ happens, and $P(B|A')$, the probability of $B$ happening over the remaining part of the sample space. Hence we must have that if $P(B|A) > P(B)$, the left over bit satisfies $P(B) > P(B|A')$. 
A: Here's one method. It uses Bayes theorem, which says that:
$$P(A|B) = \frac{P(A)}{P(B)} P(B|A)$$
for any $A$ and $B$.

Proof: You know that
$$P(B|A)>P(B)$$
Use Bayes rule, then cancel and rearrange:
$$\frac{P(B)}{P(A)} P(A|B) > P(B) \quad\Rightarrow\quad P(A|B) > P(A)$$
Since $A'$ is the complement of $A$ you have
$$1-P(A|B) < 1-P(A) \quad\Rightarrow\quad P(A'|B) < P(A')$$
Now use Bayes rule again, and cancel and rearrange one more time:
$$\frac{P(A')}{P(B)}P(B|A') < P(A') \quad\Rightarrow\quad P(B|A') < P(B)$$
Make sure you understand all of the steps and can go through them yourself.
A: Hint: $P(B)=P(B|A) P(A) + P(B|A') P(A')$, hence $P(B)$ is a convex combination of $P(B|A)$ and $P(B|A')$.
A: We are told that $\Pr(B|A)\gt \Pr(B)$. So knowing that $A$ holds makes $B$ more likely. It stands to reason that knowing that $A'$ holds makes $B$ less likely. "It stands to reason" is presumably not enough for a proof, so we give full details.
Of course we will use $\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}$. For this we need $\Pr(A)\ne 0$. From $\Pr(B|A)\gt \Pr(B)$ we get
$$\Pr(A\cap B)\gt \Pr(A)\Pr(B).\tag{$1$}$$
Note that 
$$\Pr(A\cap B)+\Pr(A'\cap B)=\Pr(B).$$
Substituting in Equation $(1)$ we get
$$\Pr(B)-\Pr(A'\cap B)\gt \Pr(A)\Pr(B),$$
which can be rewritten as 
$$\Pr(A'\cap B)\lt \Pr(B)-\Pr(A)\Pr(B)=\Pr(B)(1-\Pr(A))=\Pr(B)\Pr(A').$$
If $\Pr(A')\ne 0$, we can divide, obtaining
$$\frac{\Pr(A'\cap B)}{\Pr(A')}\lt \Pr(B).$$
This says exactly what we want, since the left side is just $\Pr(B|A')$.
