# Probability of drawing a white balls from the urn having identical white and identical black balls?

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?

• 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. Commented Feb 21, 2021 at 11:47
• The word "identical" has here no other meaning than "you draw any ball with the same probability".
– user
Commented Feb 21, 2021 at 11:51

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.