The following question has the same answers with and without replacement, but how is it possible? An urn contains 5 red, 2 white and 3 black balls.

Question

The question is as follows:

An urn contains 5 red, 2 white and 3 black balls. Three balls are drawn one-by-one, at random without replacement. Find the probability distribution of the number of white balls. Also, find the mean and the variance of the number of white balls drawn.

Using the with replacement and without replacement concepts, both give the same answer. How on earth is it possible?

I understand that this is the correct method but even without replacement, the other way too answer comes out to be 28/75.

Answer 1

both give the same answer. How on earth is it possible?

With large numbers (1000 balls, 197 yellows ....) you will probably have different mean and variance for the 2 formulas.

With small numbers (only 10 balls ...) you may be lucky (or unlucky, point of view), and obtain same mean and variance. if you consider that without replacement, mean should be less than with replacement (or the opposite), it is not consistent. When you sum all the colors, the sum of means is known to be 3 (3 balls are drawn), so you have necessarly some colors where the mean is greater in process 1 than in process 2, and other colors where it is less. And maybe some colors where the mean is the same in process 1 and process 2.

Answer 2

The probability distributions aren't the same in the two cases. Three balls are drawn and there are two white balls. So,

$$\begin{gather*} \Pr_\text{w/ repl}(\text{draw three white}) = (2/10)^3 = 1/125 > 0 \\ \text{but}, \quad \Pr_\text{w/o repl}(\text{draw three white}) = 0. \end{gather*}$$