What's wrong with my intuition of the intersection of these two events? I've got this problem I'm working on: 
Alice searches for her term paper in her filing cabinet, which has several drawers. She knows that she left her term paper in drawer $ j $ with probability $ p_j > 0 $. The drawers are so messy that even if she correctly guesses that the term paper is in drawier $ i $, the probability that she finds it is only $ d_i $. Alice searches in a particular drawer, say drawer $ i $, but the search is unsuccessful. Conditioned on this event, show that the probability that her paper is in drawer $ j $, is given by 
$ \frac{p_j}{1 - p_id_i} $ if $ j \neq i $
I defined an event A as "finding the paper in drawer j" and event B as "not finding the paper in drawer i". 
And what I'm trying to do is find $$ P(A | B) $$
I know that 
$$ P(A | B) = \frac{P(A \cap B)}{P(B)} $$
When dealing with the numerator, $ P(A \cap B) $, I ran into some confusion based on whether I evaluate $P(A \cap B) $ or $P(B \cap A)$. 
Starting with $ P(A \cap B) $, we can write this as:
$$ P(A \cap B) = P(A)P(B|A) $$ 
This can be read as the probability of finding the paper in drawer j multiplied by the probability of NOT finding it in drawer i GIVEN that its in drawer j. $ P(B|A) $ is just 1 and so we get the $ P(A) = p_j $, as does $P(A \cap B) $. 
That makes sense to me. 
But I can also look at $ P(A \cap B) $ as $ P(B \cap A) $ and if I carry out the same calculation, I don't know how to evaluate it. For example, we have 
$$ P(B \cap A) = P(B)P(A|B) $$
And this reads as the "probability of not being in drawer i multiplied by the probability of being in j given that you're not in i". In this case, $ P(A|B) $ seems to be $ p_j $, but $ P(B) $ is $ 1 - p_id_i $. 
So I get different answers if I'm evaluating $ P(A \cap B) $ vs. $ P(B \cap A) $. 
Where did I go wrong? Thanks!
 A: If $P(A|B)$ would just be $p_j$, and given that you are asked to find $P(A|B)$, you wouldn't have to use Bayes' formula! ... so that should have been a hint that something is off with your second method...
To be specific about what went wrong: $P(A)=p_j$, but $P(A|B)$ will be different from that: given that Alice searched but did not find the paper in drawer $i$ should increase the chances of finding it in drawer $j$ (unless $d_i=0$). So: unless $d_i=0$, we have that $P(A|B) > P(A)$, and so  $P(A|B) > p_j$ ... which is reflected by the formula you are asked to prove.
And by the way, I am a little confused why the formula is $\frac{p_j}{1-p_id_i}$ ... I thought it was a given that she already searched in drawer $i$. That is, I thought the question asked: given that she did not find it in drawer $i$ (but given that she did search drawer $i$), what is the chance it is in drawer $j$?  That should just be $\frac{p_j}{1-d_i}$.  The $\frac{p_j}{1-p_id_i}$ is the chance of the paper being in drawer $j$ given that the event of searching_drawer_i_and_finding_it_there does not take place ... which is also true when she doesn't search drawer $i$ at all.
Put differently: "not finding the paper in drawer $i$" is ambiguously phrased, and if you're not careful, you could interpret it in a way that is not compatible with the formula.
A: At first, note that $A\cap B$ is the same group as $B \cap A$ since this operation is commutative, therefore $P(A\cap B) = P(B \cap A)$.
Now, note that they look for the probability that the paper is in the drawer j, not that she finds it at drawer j, therefore your definition of A is not good for finding the solution.
Define $A$ - paper in drawer $j$,  $B$ - search in drawer $i$ was unsuccessful.
Therefore we get $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{1-p_id_i} = \frac{p_j}{1-p_id_i}$$ as needed.
The reason why $A \cap B = A$ is because $A \cap B^c = \emptyset$ and therefore $A \cap B = A$.
A: With the second method you end up with 
$$P(A \mid B) = \frac{P(B) P(A \mid B)}{P(B)}$$
which clearly simplifies to 
$$P(A \mid B) = P(A \mid B)$$ 
which is true, but useless. 
It's true that $P(B) = 1 - p_id_i$, but the computation for $P(A \mid B)$ is wrong. Computing $P(A \mid B)$ directly it's hard, so that's why you use the conditional formula to compute it based on $P(B \mid A)$ in the first place! 
So your second computation is wrong (other than being useless), specifically because $P(A \mid B) \neq p_j$. 
