Probability problem of A given B

Question

How can I prove that $P(A|B)\leq P(A)$? It is conceptually clear, but I want a mathematical proof.

Answer 1

It's not true as greedoid's example shows. In fact, for any given value of $P(A)\neq0,1$, the value of $P(A\mid B)$ can be anything you want by suitable choice of $B$, ranging between the two extremes of $P(A\mid A)=1$ and $P(A\mid A^c)=0$.

Answer 2

This is not true:

Say we have two fair dice.

$A$ = sum is $4$ and $B$= both show $\leq 3$.

Then $P(A|B) = {3\over 9}$ and $P(A) = {3\over 36}$