As I am currently studying for an exam on probability, I've come across a questions for which I have been unable to understand the solution.

The problem reads as follows: The are $N$ (=amount of white balls) + $M$ (=amount of black balls) balls in a box. One takes $n \le N + M$ balls out of the box. Let the random variable $X$ denote the total number of white balls removed from the box. Find the distribution of $X$.

The solution states: $$P[X = k] = \frac{\binom{N}{k}\binom{M}{n-k}}{\binom{N+M}{n}}\text{.}$$

My problem in understand this solution is: Doesn't the binomial coefficient imply that the elements are distinguishable? How can we make we make sense of - for instance - the denominator if $N + M$ means that there are indeed $N+M$ balls but only two "variations" ($N$ white ones, $M$ black ones)? Why is there no error in counting the elements this way?

Thanks in advance R.G.

  • 2
    $\begingroup$ Put ID numbers in invisible ink on the balls. That will not change the probability. $\endgroup$ – André Nicolas Jun 18 '14 at 21:43
  • $\begingroup$ Cool, thanks! That really helped me out! $\endgroup$ – R.G. Jun 18 '14 at 21:46
  • $\begingroup$ You are welcome. The main problem with trying to view them as indistinguishable is that one may inadvertently choose a sample space whose elements are not equally likely. $\endgroup$ – André Nicolas Jun 18 '14 at 23:44
  • $\begingroup$ Hey, thanks for the input! Could you maybe provide an example of such a case? The only possibility I could think of was introducing two "equal" elements in the sample space by mistake and thus ending up with an error calculating their probability. But couldn't the use of multisets help to avoid this problem? $\endgroup$ – R.G. Jun 19 '14 at 7:33
  • $\begingroup$ The classic but too simple example is flipping two identical coins simultaneously. Then the three possible outcomes are two heads, two tails, and one of each. One could believe, wrongly, that they are equally likely. I have seen this sort of mistake several times on MSE for balls in bins problems. $\endgroup$ – André Nicolas Jun 19 '14 at 7:40

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