# Calculate the probability that the $i$-th draw returns a red ball.

$$R$$, $$B$$, and $$k$$ are positive integers. An urn initially contains $$R$$ red and $$B$$ black balls. A ball is drawn at random and its colour is noted. The ball is put back in the urn along with k balls of the same colour. The process is repeated many times.

Calculate the probability that the second ball drawn is red? Calculate the probability that the $$i$$-th draw returns a red ball?

My approach Initially urn contains : $$R + B$$

\begin{aligned} P(\text{Red}) &= \frac{R}{R+B} ~\textbf{and}~ P(\text{Black}) \\ & =\frac{B}{R+B} ~\textbf{and}~ \textrm{If red is drawn then } P(\text{drawing Next Red again}) \\ &= \frac{R+k}{R+B+k}\end{aligned}

Similarly we have $$\displaystyle P(\text{drawing Next Black again}) = \frac{B+k}{R+B+k}$$

Now, \begin{aligned} P(\text{Second Red}) &= P(\text{Red} \land \text{Red}) + P(\text{Black} \land \text{Red}) \\ &= \frac{R}{R+B}\times\frac{R+k}{R+B+k} + \frac{B}{R+B}\times\frac{R}{R+B+k} \\ &= \frac{R}{R+B+k} \left(\frac{R+k}{R+B} + \frac{B}{R+B} \right) \\ &= \frac{R}{R+B}\end{aligned}

Is it correct?
And how to do the next part: probability that the $$i$$-th draw returns a red ball?

Suppose that you label the initial balls with $$0$$. For every $$n$$, when you add $$k$$ additional balls after the $$n$$-th draw, you label these new balls with the number $$n$$.
Let $$p_0 = \frac{R}{R+B}$$, and for all $$n > 0$$ let $$p_n$$ be the probability that the $$n$$-th ball drawn is red.
Suppose that the $$i$$-th ball you draw has the number $$j$$ on it (note that $$j < i$$). Then the probability that it is red is equal to $$p_j$$. With this, you can calculate the first few $$p_n$$ and form a hypothesis from there that you can prove inductively.