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I know the probability in both cases evaluate out to be 8/10. I want to know the intuition behind solving the probability of choosing a red ball when there are 8 identical red balls and 2 identical black balls. Had all balls been distinct, the result would have been obvious that there are 10 possible outcomes( since balls are distinct, each ball drawn forms a distinct outcome). Using the classical definition of probability since each outcome is equally likely the probability of choosing red ball would turn out to be number of favorable outcomes/ total number of outcomes ie 8 /10. However if 8 red balls are identical and 2 black balls are identical, the possible outcomes are only 2, either it would be a red ball or a black ball and these outcomes aren't equally likely, so we cannot apply classical definition of probability here ie total number of favorable outcomes/total number of outcomes. I want to know the intuition in such cases, about how we can relate it to the case if balls weren't identical and in what type of situations we can do so?

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  • $\begingroup$ The intuition is that the two scenarios you describe are the same -- classical probability does NOT imply that distinguishable outcomes must be equally likely. Instead, imagine that the "identical" balls might look identical to one observer but to another observer with better vision, they are distinguishable ... yet the experiment with red/black outcomes has the same distribution to both of them. $\endgroup$ – Ned Apr 18 at 19:49
  • $\begingroup$ @Ned But when I consider case 2 it is explicitly mentioned the balls are identical, so if we do not consider this- "Instead, imagine that the "identical" balls might look identical to one observer but to another observer with better vision, they are distinguishable" then the outcome we can see is either a red ball or a black ball. We cannot distiguish among same colour balls. So when I take this scenario, my 2 outcomes are a red ball or black ball, so how will I take probability in this case. $\endgroup$ – Amisha Bansal Apr 19 at 6:25
  • $\begingroup$ @ Amisha Bansai You count the number of each and proceed in the obvious way to get the distribution P(red) =0.8, P(black) = 0.2 (if your model is "a ball is chosen uniformly"). At some level the red balls ARE distinguishable (you have 8 of them) even if you can't tell them apart. Is your question philosophical -- then I have nothing useful to say -- or mathematical, in which case the uniform distribution on balls -- not colors -- means the color distribution is 0.8 and 0.2. What's the problem? 2 outcomes doesn't mean 50/50 probability!! $\endgroup$ – Ned Apr 19 at 12:00
  • $\begingroup$ math.stackexchange.com/questions/995519/… : this link should help $\endgroup$ – Shubham Gupta Apr 20 at 15:23

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