Drawing a ball of particular color at the i'th turn from an urn I've an urn with $r$ red and $w$ white balls. I'm randomly drawing a ball from the urn each round(without replacement). I'm asked the probability of the event $R_i$:Drawing a red ball from the urn in i'th turn. 
$R_1$ is simply $\frac{r}{r+w}$.
$R_2$ can be calculated as follows: $$ P(R_2) = P(R2|R1)P(R1) + P(R2|R1^c)P(R1^c) $$
$$=\frac{r-1}{r+w-1}\cdot\frac{r}{r+w} + \frac{r}{r+w-1}\cdot\frac{w}{r+w}  $$ 
Since we're dealing with conditional probability, the formula for the $i$'th turn would be very messy as far as I can see. I feel there's a simple answer to this question but I'm missing the intuition. So, how should I approach to this problem?
Further, I'm asked $P(R_5|R_3)$. Calculating this also seems very cumbersome. Is there an easy way for it too ?

You can ignore all the balls you don't look at.  $P(R_2)$ is $\frac r{r+w}$ if you don't look at the first ball.  The expression you have will simplify to this. 
 Imagine drawing two balls, then just before you look at the second one you switch them.  Similarly, when asked for $P(R_5|R_3)$ you can ignore the first, second, and fourth draws.  You are told the third ball is red, so when you draw the fifth you have $r-1$ red balls and $w$ white balls, giving a probability of $\frac {r-1}{w+r-1}$
