$R$, $B$, 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?
Best Answer
Your approach is correct.
As for how to prove the general case, consider the following trick:
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.