Alison 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 $p_i \ge 0$. The drawers are so messy that even if she correctly guesses that the term paper is in drawer i, the probability that she finds it is only $d_i$. Alison 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 $\frac{p_j}{( 1 - p_i d_i )}$ if $j \ne i$ and $\frac{p_i(1-d_i)}{( 1 - p_i d_i )}$ if $j = i$
If $A \equiv$ Event Alison's search of drawer i is unsuccessful
$B_i \equiv$ Event the term paper is in drawer i
$P(A\cap B_i)=P(A|B_i)P(B_i)=p_i(1-d_i) $ where $P(A|B_i)=1-di$ because the problem says that given she identifies the right drawer the probability she retrieves the term paper is $d_i \equiv P(A^c|B_i)$ and I'm assuming the event of the term paper being in the drawer can be taken as the event of Alison having the knowledge of her having put the paper in drawer i as $p_i$.
$P(B_i|A)=\frac{P(B_i\cap A)}{P(A)}=\frac{P(A|B_i)P(B_i)}{P(A)}$
Let $C \equiv$ Event that Alison identifies the right drawer
Probability of successful search = $P(C \cap B_i)=P(C)P(B_i)=p_id_i$(Is this correct?)
Probability of unsuccesful search=$1-p_id_i \equiv P(A)$
which brings $P(B_i|A)= \frac{p_i(1-d_i)}{1-p_id_i}$
and if $i \ne jP(B_j|A)=\frac{ P(B_j \cap A)}{P(A)}$ and once again assuming the event of the being in drawer j is the same as the event of Alison having the knowledge of having put her term paper in drawer j as $p_j$ we can say that $P(B_j \cap A)=P(B_j)=p_j$ which follows the result.
Are my events identified correctly? and Is my reasoning correct? I tend to get confused between $P(A|B)$ and $P(A\cap B)$ not conceptually but from the point of solving problems.How do I make out the difference?
You get the right results, but not sure about the reasoning, since it's not clear that $C$ and $B_i$ are independent events.
Your event $C$, as written, is vague. I could interpret "identifying the right drawer" as equivalent to $B_i$. The way you use it, it really is more like this:
$$P(C) = P(\neg A) = 1-P(A) = 1 - [P(A|B_i)P(B_i) + P(A|\neg B_i)(1-P(B_i)] $$
I got the terms in brackets by using a common tactic in conditional probability calculations: the law of total probability; $B_i, \neg B_i$ partition the outcome space so we can get $P(A)$ using only conditional probabilities and the probabilities of each partition.
We know $P(A|\neg B_i) = 1$, so substituting in the known probabilities we get:
$$1 - [(1-d_i)p_i + 1\cdot(1-p_i)]=1 - [p_i-p_id_i + 1-p_i] = p_id_i \;\;\square$$
Deriving the expected results
For clarity, let's rename $A$ to $A_i$ to indicate an unsuccessful search of drawer $i$. With our knowledge of $P(A_i)$ the application of conditional reasoning is straightforward:
$$P(B_i|A_i) = \frac{P(A_i|B_i)P(B_i)}{P(A_i)} = \frac{p_i(1-d_i)}{1-p_id_i}$$
And
$$P(B_j|A_i) = \frac{P(A_i|B_j)P(B_j)}{P(A_i)} = \frac{1\cdot P(B_j)}{P(A_i)} = \frac{p_j}{1-p_id_i}$$
Intuitively this makes sense too
We can assume that $p_i \in (0,1)$ since $p_i = 1$ makes the problem trivial and $p_i=0$ makes choosing $i$ irrational. Therefore, $0< 1-p_id_i < 1$, which means
$$\frac{p_j}{1-p_id_i} > p_j$$
So $A_i$ increases the posterior probability that the paper is in one of the other drawers.
Conversely, we would expect that the probability of the paper being in drawer $i$ to decrease:
$$ d_i > p_id_i \implies \frac{1-d_i}{1-p_id_i} < 1 \implies \frac{p_i(1-d_i)}{1-p_id_i} < p_i$$