A box contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white.
I found the answer= m/(m+n). But I cannot understand one thing. Why is it independent of k?
Let's take the first draw. The expected ratio is
$$\frac{m}{m+n} \times \frac{m+k}{m+n+k} + \frac{n}{m+n} \times \frac{m}{m+n+k} = \frac{m}{m+n}$$
and something similar happens in future: each round (without knowing the results of previous rounds) you have a probability $\frac{m}{m+n}$ of adding $k$ white balls and a probability $\frac{n}{m+n}$ of adding $k$ black balls, so the expected proportion of white balls remains a constant $\frac{m}{m+n}$ after that round and so by induction after every round.