Probability of getting red ball at ith step 
An urn has r red and w white balls that are randomly removed one at a
  time. Let $R_i$ be the event that the $i$th ball removed is red. Find
  $P(R_i)$

I started with calculating $P(R_1) = \frac{r}{r+w}$. Next, 
$$P(R_2) = P(R_2|R_1)\cdot P(R_1) + P(R_2|W_1)\cdot P(W_1) = $$
$$ = \frac{r-1}{r-1+w}\cdot\frac{r}{r+w}+\frac{r}{r+w-1}\cdot\frac{w}{r+w} = \frac{\left(r-1+w\right)r}{\left(r+w-1\right)\left(r+w\right)} = \frac{r}{r+w}$$
And so on. The answer is $P(R_i) = \frac{r}{r+w}$.
The textbook's answer shows that the calculations weren't necessary:

$\frac{r}{r+w}$ because each of the $r + w$ balls is equally likely to
  be the $i$th ball removed.

I don't see why each ball is equally likely. At any ith step we don't have $(i-1)$ balls! Could someone help me to understand this?

All orderings of the balls are equally likely. So the probability that the $i$-th ball is red is the same as the probability that the first ball is red. 
Think of the $n=r+w$ balls as people. The probability that Charlie is the $i$-th person chosen is the same as anyone else's. So the probability that Charlie is the $i$-th person chosen is $\frac{1}{n}$.
