Probability of Getting a Red Ball I have a simple and straightforward question.

A box contains $n$ balls, of which $r$ are red ($r$ and  $n$ are both positive integers, and  $r \leq n$; suppose further that $n$ is even). Consider what happens when the balls are drawn from the box one at a time, at random without replacement. Determine:
$\quad$ (a) The probability that the first ball drawn will be red;
$\quad$ (b) The probability that the $\left(\frac{n}{2}\right)^{\text{th}}$ ball drawn will be red;
$\quad$ (c) the probability that the last ball drawn will be red.

I'm not sure how to approach questions (b) and (c). I understand that (a) is $\frac rn$ because the probability of the very first ball being red is the ratio of all the red balls over the total number of balls, but I don't know how to extend this idea to the $i^{\, \text{th}}$ ball.
Thank you for your time.
 A: Let me try a different approach.
For an experiment, let $r=1,$ that is, suppose there is just one red ball.
We now pull all of the balls out one at a time without replacement.
Is the red ball more likely to be the first out? The second? The last? The one at the halfway mark?
You can actually work out the probabilities using conditional probability if you like,
or you can imagine the process of pulling out balls as a process of arranging the balls randomly in a line from first to last. Either way, you should find that the red ball has exactly a $\frac 1n$ probability to show up at any particular time:
$\frac1n$ to be the first ball, $\frac1n$ to be the second,  $\frac1n$ to be the last, $\frac1n$ to be the one at the halfway mark.
In short, the $k$th ball drawn has $\frac1n$ probability to be the red one.
Experiment over.
Now a new experiment.
Suppose we have two red balls, but for convenience let's write the number $1$ on one of them and $2$ on the other. Now consider the $k$th ball drawn, $1\leq k\leq n.$
This ball has $\frac1n$ probability to be red ball number $1$
(for the same reason as in the first experiment)
and $\frac1n$ probability to be red ball number $2$.
These two events are mutually exclusive, so the probability that at least one of them happens is the sum of their probabilities, $\frac2n.$
Experiment over.
Now back to the original question, for arbitrary $r$.
Put the numbers $1,\ldots,r$ on the red balls, every red ball with a different number. Each of those balls has a $\frac1n$ probability to be the $k$th ball drawn.
There are $r$ possible events in which the $k$th ball drawn is one of those balls, each with probability $\frac1n$, each mutually exclusive with each other so that the probability of at least one of those events happening is the sum of their individual probabilities.
A: I regard this as an unfair question, because probability questions typically require the student to have first developed his intuition.
This is a trick question:  The answer to all three questions [2a, 2b, and 2c] is the same.
The easiest way to see this is to pretend that question 2c alone had been posed.  I would reason as follows: given no other information, the chance of the last ball drawn being red is no different than the chance of the first ball drawn being red.
Edit 
A simpler form of the problem would have been if the balls are drawn with replacement.  In this simpler form, it would have been easier to see that the answer to all three questions is the same.
The problem's flavor stems from the answers being the same regardless of whether the balls are drawn with or without replacement.  This is unusual; in most probability problems, it matters whether the balls are drawn with or without replacement.
A: 
"...but I don't know how to extend this idea to the "ith" ball..."


Fix some $i\in\{1,2,\dots,n\}$.
We want to find the probability that the ball that is drawn as $i$-th ball is a red ball.
All $n$ balls are candidates for becoming the $i$-th ball drawn and each of them has equal probability to become this ball.
(Especially this fact must land deep in your mind and will enrich your intuition for probability)
Well, $r$ of these balls are red, so the probability that the $i$-th ball drawn is red equals $\frac{r}{n}$.

We could make it a bit more formal like this:
Give the red balls numbers $1,2,\dots,r$ and the non-red balls the numbers $r+1,\dots,n$.
Let $E_k$ denote the event that the ball with number $k$ is the $i$-th ball drawn.
Then $P(E_k)$ does not depend on $k$ and $\sum_{k=1}^nP(E_k)=1$.
This allows us to conclude that $P(E_k)=\frac1{n}$ for every $k\in\{1,\dots,n\}$.
The event that a red ball is drawn as $i$-th ball is $\bigcup_{k=1}^rE_k$ so that:$$P(\text{ a red ball is drawn as }i\text{-th ball})=P\left(\bigcup_{k=1}^rE_k\right)=\sum_{k=1}^rP(E_k)=\frac{r}{n}$$

This works for every $i\in\{1,\dots,n\}$ and especially for $i=1$, $i=\binom{n}2$ and $i=n$ (corresponding with 2(a),2(b),2(c)).
