Question

Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in drawer j with probability pj>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 di. 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
pj/(1−pidi) if j not equals to i
pi(1−di)/(1−pjdi) ifj=i.

Answer 1

You made a mistake with the probability $p_{j}$, which should be $p_{i}$ in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer $j$, is given by:

(1) $\frac{p_{j} }{1-d_{i}p_{i} } , if j \neq i,$ and

(2) $\frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} } , if j = i.$

so we can say:

$A$ is the event that you search drawer $i$ and find nothing,

$B$ is the event that you search drawer $i$ and find the paper,

$C_{k}$  is the event that the paper is in drawer $k, k = 1, ..., n.$

this gives us:

$P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}$

$P(A) = 1 - P(B) = 1 - d_{i} p_{i}$

Solution to Part (1):

if $j \neq i$, then $P(A \cap C_{j} ) = P(C_{j} )$,

this means that

$P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)} = \frac{P(C_{j} )}{P(A)} = \frac{p_{j} }{1-d_{i}p_{i} }$

as needed so part one is solved.

Solution to Part(2):

so we have now that if $j$ = $i$, we get that:

$P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}$

remember that:

$P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}$

this implies that:

$P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )$

so we just need to combine the above relations to get:

$P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} }$

as needed so part two is solved.