Probability of taking balls without replacement from a bag question

You are definitely on track. In the first probability, order does not matter. You have drawn $n-1$ balls, and one of them is red. The total number of outcomes is the total number of ways to draw $n-1$ balls from $N$. The number of outcomes you are interested in are given by:

$$\dbinom{2}{1}\dbinom{N-2}{n-2}$$

So, the probability that you have drawn exactly one red ball in the first $n-1$ draws is:

$$\dfrac{\dbinom{2}{1}\dbinom{N-2}{n-2}}{\dbinom{N}{n-1}}$$

Now, multiply by the probability that the last ball is red (you were correct):

$$\dfrac{1}{N-(n-1)}$$

Final probability:

$$\dfrac{2(n-1)}{N(N-1)}$$

Here's another approach, depicted graphically below for $N = 8$.

We can construct a grid, where the equiprobable slots are represented by squares, indexed by the sequence position of the two red balls. The X's down the diagonal indicate that the two red balls cannot occupy the same sequence position.

The result of the experiment is the maximum of the two positions; for instance, the cases where $n = 5$ are marked by a red $5$. From this it can be seen that there are $2n-2$ different squares where the number of balls drawn is $n$, out of a total of $N^2-N$ possible squares. Thus, the desired probability is

$$ \frac{2n-2}{N^2-N} = \frac{2(n-1)}{N(N-1)} $$

as in InterstellarProbe's answer.