Probability of finding a letter Imagine that you are searching for an important letter that you received some time ago. Usually your mother puts your letters in the drawers of your desk after you have read them. She remembers to do this in $p$$\%$ of the cases , and in $(100-p)$$\%$ of the cases she leaves them somewhere else. $(100>p>0).$ 
There are $n(≥2)$ drawers $\big(\frac{100n}{p}is \space not \space an \space integer\big)$ , in your desk. If indeed your mother has placed the letter in your desk , you know from past experience that it is equally likely to be in any of the $n$ drawers. You start a thorough and systematic search of the drawers of your desk ;
(a) if you don't find the letter in the first $j\space(n-1≥j≥1)$ drawers , what is the probability that the letter is in the desk i.e. in one of the remaining $n-j$ drawers ?
(b) if you don't find the letter in the first $j$ drawers , what is the probability that the letter is in the next drawer ?   
[ if $\frac{100n}{p}$ is an integer then I can solve the problem by changing it to a more intuitively clear version,isomorphic to the original problem ; but I can't do so in case $\frac{100n}{p}is \space not \space an \space integer$ , that's why I have mentioned it.]   
 A: Divisibility issues are of no importance. For the first problem, we want the probability the letter is in the desk given that the search in the first $j$ drawers has failed.
As has been pointed out by Thomas Andrews, your $p$ is a number between $0$ and $100$. This is a very non-standard usage. I will try to deal with it by defining the number $p^\ast$ to be your $p$ divided by $100$.
Let $F$ be the event the search of the $j$ drawers has failed, and $D$ the event the letter is in the desk. We want $\Pr(D|F)$. By the defining formula for conditional probabilities, we have 
$$\Pr(D|F)=\frac{\Pr(D\cap F)}{\Pr(F)}.$$
The probability of $D\cap F$ is $p^\ast\cdot \frac{j}{n}$.  
For the probability of $F$, note that the search fails if (i) the letter is in the desk and the search fails, or (ii) if the letter is not in the desk. We already calculated the probability of (i). The probability of (ii) is $1-p^\ast$, so the probability of $F$ is $p^\ast\cdot\frac{j}{n}+1-p^\ast$. Now you have all the ingredients.
For (b), we can use the result of (a). Given the search has failed, we know the probability the letter is in the desk. Given that the letter is in the desk, and $j\lt n$, the probability it is in the first unsearched drawer is $\frac{1}{n-j}$. So we multiply the answer to (a) by $\frac{1}{n-j}$.
A: Let $m=100n/p$ be an integer . Without loss of generality , let us suppose that there are $m$ drawers in your desk , your mother "always" puts your letters in the drawers of your desk after you have read them.You know that the letter is equally likely to be in any of the $m$ drawers. You notice , however , that drawers #$(n+1)$ , #$(n+2)$ ,...,#$m$ are locked,and your    has gone outside with the keys; you have first $n$ drawers to search for. you realize that there is a $100n/m=p$$\%$ chance  that the letter is in one of the unlocked drawers. You start a thorough and systematic search of the $n$ unlocked drawers. All the questions are exactly as before,in (a) you are asked about the probability that the letter is one of the remaining $(n-j)$ unlocked drawers and in (b) about the probability that it is in the next drawer given it is not in the first $j$ drawers. Now considering this new(isomorphic) version of the problem , it is easy to see that the required probability for question (a) is $\frac{n-j}{m-j}=\frac{n-j}{\frac{1oon}p-j}$ and for question (b) it is $\frac1{m-j}=\frac1{\frac{100n}p-j}$ . I think these are not in accordance with your solution AndréNicolas; but I'm quite sure that this solution is correct I obtained this by following a solution of a problem in a probability book by Ruma Falk , there $p=80 , n=8 $ were given and was solved by same method.
