A simple question in probability that somehow got me confused. Let $(X_1,...,X_N)$ be a random arrangement of the numbers $1,...,N$. We say that the $n-th$ component is inferior, if there is no $i<n$ such that $X_n>X_i$. 
a. What is the probability that the $n-th$ component is inferior?
This one is clear, as the answer states: In a set of $n$ components, $X_1,...,X_n$ there is one minimum. $X_n$ is the minimum with probability $1\over n$. (Is that correct? I am not sure I got it well.)
b. For $m<n$, find that probability that the $n-th$ component is inferior, assuming the $m-th$ one isn't. 
I have a problem here: the answers say that it stays $1\over n$ because there is no independence(it didn't explain much.). I, however, arrived at $1\over n-1$ for if $X_m$ is not inferior, then it is not the minimum, and the probability that $X_n$ is the minimum is the number of minimums(1) divided by the number of components that could be that minimum. What am I missing here? 
 A: Both of the answers you quote are right.
For the first part, consider the first $n$ numbers. Ignore their concrete values; just note that they're $n$ different numbers than can be ranked from highest to lowest. It doesn't matter which ones they are and how many or which gaps are in between their actual values – what matters is that all $n!$ permutations of them are equally like, so the chance of the least being the last is $1$ in $n$.
In the second case, your argument seems plausible at first sight – the $m$-th can no longer be the minimum, so that should raise the probability of the $n$-th being the minimum. But look at it this way: First I tell you that these particular $n$ numbers landed in the first $n$ slots. As above, that has no bearing on the probability of the $n$-th number being inferior, because that doesn't depend on the concrete values and only on their order. Now I tell you that the $m$-th number isn't the minimum of the first $m$ numbers. By saying that, I've eliminated $1$ out of $m$ of the $m!$ permutations of the first $m$ numbers. But again that has no bearing on whether they're higher or lower than the $n$-th number – it's just a piece of information on their order. So the probability that the $n$-th component is inferior hasn't changed and is still $1/n$ under this condition.
