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In an urn there are $m$ white balls and $n$ black balls. If a ball is drawn from the urn, then what is the probability that the drawn ball is white, given the condition that all white balls are identical and all black balls are identical?

My approach: Number of cases favorable to the event is $m$, and total number of cases for the sample space is $m+n$. So, $$ P(\text{white ball}) = \frac{m}{m+n} $$

But question is given that 'all white balls are identical and all black balls are identical'.

In the case of identical balls, number of cases favorable to the event is $1$ (no of ways of selection of one white ball from the $m$ identical white balls).

And total number of cases for the sample space are $2$ cases (either white ball is selected or black ball is selected). Here, $$ P(\text{white ball}) = \frac{1}{2} $$

I know that this $\frac{1}{2}$ is incorrect. Could any one help me out?

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    $\begingroup$ What matters here is how many of the balls are white. To see the fallacy in reducing the problem to two possible outcomes, suppose $m = 1$ and $n = 99$. Clearly, in my example, you are not equally likely to select a white ball or a black ball. $\endgroup$
    – N. F. Taussig
    Commented Feb 21, 2021 at 11:47
  • $\begingroup$ The word "identical" has here no other meaning than "you draw any ball with the same probability". $\endgroup$
    – user
    Commented Feb 21, 2021 at 11:51

2 Answers 2

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There are $m + n$ equally likely outcomes of drawing one ball. The number of outcomes that consist of a white ball is $m$. Thus the probability is $$ P(\text{white ball}) = \frac{m}{m+n}, $$ as you suggested initially. The "identical" modifier is just to suggest that when you count a "white ball", there's no way to distinguish between different white balls.

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It is just a matter of what you mean by "probability". Surely you can assign 0.5 to both cases, but that is not natural(telling us nothing interesting in real life), which means not work with our instinct. So the most natuaral way of defining probability is just assigning equal probability to every "possible cases".

Now I think this is the subtle point here that you are confused of. "Balls are identical" doesn't mean there is only one ball, it means you cannot tell the difference between white balls--the only factor that counts is the color. See? Identical here only justifies our intuition that we SHOULD assign equal probability to all balls, since only color matters.

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