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 \begin{cases} \cfrac{p_j}{1-p_id_i} & \text{if $j \neq i$},\\ \cfrac{p_i(1-d_i)}{1-p_id_i} & \text{if $j = i$} \end{cases}
I followed that approach, and I couldn't find anything.
Sorry for grammatical mistakes.
Let $F$ be the event that she fails to find the paper in drawer $i$, and $I$ be the event that the paper is in drawer $I$.
Then $p(F|I)=p_i(1-d_i)$ and $p(F|\text{not }I)=1-p_i$. Hence $$p(I|F)=\frac{p(F|I)}{p(F|I)+p(F|\text{not }I)}=\frac{p_i(1-d_i)}{p_i(1-d_i)+(1-p_i)}=\frac{p_i(1-d_i)}{1-p_id_i}$$ So the prob that it is not in drawer $i$ after her search of that drawer is $$1-\frac{p_i(1-d_i)}{1-p_id_i}=\frac{1-p_i}{1-p_id_i}$$ Clearly her search of drawer $i$ does not affect the ratios of the other probabilities to each other, so the probability of finding it in drawer $j\ne i$ becomes $$\frac{p_j}{1-p_id_i}$$