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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?

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    If you show all steps, maybe the $k$ just cancels. – 2017-02-20
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    Here's one way to think about it: The expected number of white and black balls added is proportional to the original ratio of white and black balls; therefore, on average the draw-and-add process does not change the distribution of white and black balls. – 2017-02-20
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    But you only add k balls of a single colour. So the ratio doesn't remain the same. – 2017-02-20
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    It's true that k cancels but is there an intuitive explanation for the cancellation? – 2017-02-20

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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.