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Let there be $100$ balls in an urn out of which $50$ are red and $50$ are green. Let $A$ be the event of drawing $75$ balls from the urn in which $n$ balls are red in colour, where $25 \leq n \leq 50.$ Find $\Bbb P(A).$

Note $:$ All the red (resp. green) balls are indistinguishable.

I know the result if the balls are distinguishable in which case $$\Bbb P(A) = \frac {\binom {50} {n} × \binom {50} {75-n}} {\binom {100} {75}}.$$

How can I solve this question for indistinguishable balls? Any help will be appreciated.

Thanks in advance.

Anil Bagchi.
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    Your result for distinguishable balls is incorrect. The probability does not change if you number the balls so that you can distinguish them, it's still given by the formula I gave here (https://math.stackexchange.com/questions/3717811/what-is-bbb-ex/3717822?noredirect=1#comment7641398_3717822). The difference only comes into play during the computation of said probability, but the end result is the same. – Ant Jun 13 '20 at 12:09
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    @Ant please avoid answering this question as I am not convinced to the answer related to it you have given before. Let it be answered by other users in this site. – Anil Bagchi. Jun 13 '20 at 12:17
  • @Ant the formula given here coincides with yours (easily confirmed either numerically or algebraically). Both are correct. – lulu Jun 13 '20 at 13:16
  • @Phibetakappa I understood that, that's why I only posted a comment and did not answer. Apologies if I have confused you, I didn't realise your formula was correct as well. – Ant Jun 13 '20 at 15:37

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There is no distinction between distinguishable and indistinguishable balls in probability issues. Saying "indistinguishabe" you just mean that any distinction between the balls is ignored. It becomes obvious if you number the balls. Obviously the probability to draw $n$ red balls does not depend on the fact whether the balls are numbered or not.

It can be different for really indistinguishable objects such as quantum particles. This is the reason why they do not obey the classical Boltzmann statistics. But for classical objects such as balls this plays no role.

user
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  • Then where am I doing mistake in computing $\Bbb P(A)$? – Anil Bagchi. Jun 13 '20 at 13:15
  • Did I say that you made a mistake? – user Jun 13 '20 at 13:16
  • Out of $75$ balls I need to draw $n$ red balls which can be taken from $50$ red balls in $\binom {50} {n}$ ways. Then all the remaining $(75-n)$ balls have to be green in colour. So this can be taken from $50$ green balls in $\binom {50} {75-n}$ ways. – Anil Bagchi. Jun 13 '20 at 13:21
  • But why did ant mention in his comment that I made mistake @user? – Anil Bagchi. Jun 13 '20 at 13:22
  • Yes, I understand the background of the formula given in your question. I do not answer the comment of @ant, I answer your question. But if you need, I can confirm that your formula is correct. – user Jun 13 '20 at 13:23
  • It's ok. But didn't he (ant) mislead me by saying that my result is incorrect? – Anil Bagchi. Jun 13 '20 at 13:25
  • It will be better if you confirm my answer to clear all my doubts. I also eager to know that why do the concepts "indistinguishability" or "distinguishability" is of no importance here. I want a rigorous canonical answer to this question. Thanks for your patience. – Anil Bagchi. Jun 13 '20 at 13:30
  • @Phibetakappa I confirm your answer. – user Jun 15 '20 at 08:12