An urn contains $m$ green and $n$ red balls. $K (< m, n)$ balls are drawn and laid aside, their colour being ignored.
Then one more ball is drawn. Then the probability that it is green.
What I don't understand is that how do we write favorable cases when one more ball is drawn as color of $K$ balls was ignored in first draw.
Answer given is $\frac{m}{m+n}$
Best Answer
A green ball is equally likely to appear anywhere in the sequence. Since $m$ of the $m + n$ balls are green, the $(k + 1)$st ball selected has probability $$\frac{m}{m + n}$$ of being green. To see this, imagine lining up the $m + n$ balls in some order in the dark, then reaching for the ball in the $(k + 1)$st position.