Conditional probability : can P(A|B) > P(A)? I just started to learn about probabilities and I am doing some simple exercises. But I'm having problems solving and even understanding an exercise.
We have a partition $B_1, B_2 ... B_k$ of event space $\Omega$ with $B_i \cap B_j = \emptyset $
Let A be an event such that $P(A) > 0$.
Prove that if: $P(B_1|A) < P(B_1)$ then $P(B_i|A) > P(B_i)$ for at least one value of $i$ from ${2,3..k}$
I know it's a very easy exercise but I can't manage to see how can $P(B_i|A) > P(B_i)$
Generally speaking, how can $P(A|B) > P(A)$ ? $P(A|B)$ means the probability of A assuming that B has occurred. So isn't it supposed to be <= P(A) (with equality when A and B are independent events)
 A: Hint:
$$
P(B_1|A)+\sum_{i=2}^kP(B_i|A)=1=P(B_1)+\sum_{i=2}^kP(B_i).
$$
As for $P(B_i|A)>P(B_i)$, note that 
$$
P(B_i|A)=\frac{P(B_i\cap A)}{P(A)}.
$$
While it is true that $P(B_i\cap A)\leq P(B)$, we also have $P(A)\leq 1$, so it's entirely possible that the fraction $\frac{P(B_i\cap A)}{P(A)}$ is larger than $P(B_i)$.
I recommend that you make up a numerical example for the joint distribution for $\{X,Y\}$ where $X\in\{1,2\}$ and $Y\in\{1,2\}$. Let $A$ be the event $X=1$ and $B_1$ and $B_2$, respectively, be the events $Y=1$ and $Y=2$. Then, because
$$
P(B_1|A)+P(B_2|A)=P(B_1)+P(B_2)
$$
so, twitching some numbers around if necessary, you will have either
$$
P(B_1|A)>P(B_1)\quad\text{or}\quad P(B_2|A)>P(B_2).
$$
A: Another idea: Assume that $P(B_i|A)\leq P(B_i)$, for all $i=2,\ldots,k$. By the law of total probability: \begin{align*}P(A)=\sum \limits_{i=1}^k P(A|B_i)P(B_i)=\sum \limits_{i=1}^k P(B_i|A)P(A)&< P(B_1)P(A) + \sum \limits_{i=2}^k P(B_i|A)P(A)\\ & \leq P(B_1)P(A)+\sum \limits_{i=2}^k P(B_i)P(A)
\\& =P(A)\sum \limits_{i=1}^k P(B_i)=P(A),
\end{align*}
which is a contradiction.
