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.

  • $\begingroup$ If the 50 chosen include $w$ white, then they include $50-w$ black, and the remaining 50 are $50-w$ white and $w$ black, so after the replacement you have $100-2w$ white and $2w$ black. So you need to compute, for all values of $w$ that makes both $100-2w$ and $2w$ at least 40, the probability that 50 drawn at random have $w$ white. Then add. $\endgroup$ – Gerry Myerson Jul 9 '13 at 9:03
  • $\begingroup$ I can with certainty say that the probability of 59-41, 57-43, 55-45, 53-47 , ..., 41-59 occurrence are all zero! $\endgroup$ – Ali Jul 9 '13 at 9:06
  • $\begingroup$ $\frac{35973547957477625884325468}{36983630698520598729769977} \simeq 0.972688$ $\endgroup$ – Ali Jul 9 '13 at 9:22

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: $$\sum\limits_{k=20}^{30}\frac{\binom{50}{k}^2}{\binom{100}{50}}\approx 97\%$$

  • $\begingroup$ A little slip: you need $20\le k \le 30$ $\endgroup$ – Empy2 Jul 9 '13 at 9:18
  • $\begingroup$ Oops, thank you. It should be symmetrical after all. $\endgroup$ – Tomas Jul 9 '13 at 9:19
  • $\begingroup$ Leaving OP nothing to do. $\endgroup$ – Gerry Myerson Jul 9 '13 at 13:28
  • $\begingroup$ @GerryMyerson Not sure about that. If the batch consisted of only $10$ balls then apparently $\sum\limits_{k=0}^{10}\frac{\binom{50}{k}^2}{\binom{100}{10}}>1$ . What is the problem? $\endgroup$ – Pranasas Jul 10 '13 at 13:03
  • $\begingroup$ The problem is, you asked a question, and Tomas deprived you of the joy of working out even a small part of the solution by doing it all for you. Then again, it may be that in the light of your further question you don't actually understand what Tomas has done, so maybe there is still hope that you have something to enjoy working on. $\endgroup$ – Gerry Myerson Jul 10 '13 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.