There are $100$ balls in the urn – $50$ black and $50$ white. A person randomly chooses a batch of $50$ balls and replaces it with a batch of inversely colored balls (i.e. for example a batch of $34$ black and $16$ white balls would be replaced by a batch of $16$ black and $34$ white balls).
It is obvious that after replacement took place there could be anything from all black to all white balls. But what is the probability that one can still find at least $40$ balls of either color?
Honestly I'm not even sure how to figure out the probability of urn still having 50-50 balls, and for my original question apparently I have to sum the probabilities of $60-40$, $59-41$, $58-42$, $\ldots$, $41-59$, $40-60$ ball combinations.
Best Answer
If the person chooses $k$ black and $50-k$ white balls, then after inverted replacement, we have $100-2k$ black and $2k$ white balls.
You will find at least $40$ balls, iff $2k\geq 40$ or $100-2k\geq 40$, in other words, iff $20\leq k\leq 30$.
There are $\binom{100}{50}$ possible batches. Now count the batches with exactly $k$ black balls. You can choose the $k$ black balls in $\binom{50}{k}$ ways and the $50-k$ white balls in $\binom{50}{50-k}=\binom{50}{k}$, ways. So the result should be: $$\sum\limits_{k=20}^{30}\frac{\binom{50}{k}^2}{\binom{100}{50}}\approx 97\%$$