# Pólya's urn scheme, proof using conditional probability and induction

Problem

An urn contains $B$ blue balls and $R$ red balls. Suppose that one extracts successively $n$ balls at random such that when a ball is chosen, it is returned to the urn again along with $c$ extra balls of the same color. For each $n \in \mathbb N$, we define $R_n=\{\text{the n-th ball extracted is red}\}$, and $B_n=\{\text{the n-th ball extracted is blue}\}.$

Prove that $P(R_n)=\dfrac{R}{R+B}$.

I thought of trying to condition the event $R_n$ to another event in order to use induction. For example, if $n=2$, I can express $$P(R_2)=P(R_2|R_1)P(R_1)+P(R_2|B_1)P(B_1)$$$$=\dfrac{R+c}{R+B+c}\dfrac{R}{R+B}+\dfrac{R}{R+B+c}\dfrac{B}{R+B}$$$$=\dfrac{R}{R+B}.$$

Now, suppose the formula is true for $n$, I want to show it is true for $n+1$.