Conditional probability - Proof I interpret this as the statement that if A is true then B is more likely.
I have
$P(B \mid A) > P(B)$
Now I want to formulate and prove:


*

*If not-B is true, then A becomes less likely

*If not-A is true, then B becomes less likely
If I get help with the first one I am sure I will be able to do the second one myself. 
Thanks!

What we want to show for the first is $P(A)>P(A|B^c)$. Let's start by using the law of total probability, as suggested by Alex. The law states that
$$P(A) = P(B^c)P(A|B^c)+P(B)P(A|B).$$
Now, we can solve for $P(A|B^c)$ and we get
$$P(A|B^c) = \frac{P(A)-P(B)P(A|B)}{P(B^c)}.$$
Noting that $P(B)P(A|B) = P(A \cap B) = P(A)P(B|A)$, we get
$$P(A|B^c) = \frac{P(A)-P(A)P(B|A)}{P(B^c)}.$$
Since we assumed that $P(B|A) > P(B)$ and using $P(B^c)=1-P(B)$, we get the inequality
$$P(A|B^c) = \frac{P(A)-P(A)P(B|A)}{P(B^c)} < \frac{P(A)-P(A)P(B)}{1-P(B)} = P(A)\frac{1-P(B)}{1-P(B)} = P(A).$$
For the second statement, just follow similar steps and you should get to an answer.
