Problem:
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 drawer $i$, 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 $\displaystyle{\frac{p_i(1 - d_i)}{1 - p_id_i}}$ if $j = i$.
Solution:
Let $A$ be the event that Alice does not find her paper in drawer $i$. Since the paper is in drawer $i$ with probability $p_i$, and her search is successful with probability $d_i$, the multiplication rule yields $P(A^c) = p_i d_i$, so that $P(A) = 1 − p_i d_i$. Let $B$ be the event that the paper is in the drawer $j$. If $i = j$, then $\displaystyle{P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(B)P(A|B)}{P(A)} = \frac{p_i(1 - d_i)}{1 - p_id_i}.}$
My question:
I'd like to get a better idea of how they used the multiplication rule. Let $E_1$ be the event that her paper is in $i$ and $E_2$ - the event she finds her paper. Then $P(A^c) = P(\{E_1 \cap E_2\}) = P(E_1)P(E_2) = p_id_i$ because the events $E_1$ and $E_2$ are independent. Does that make sense? Thanks.
Yes, that is right. For independent events, simply multiply the probabilities of the events to find the probability of both ocurring.