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Suppose we have a box, with only one small hole. Suppose 10 distinct black balls and 20 distinct white balls are put in the box. Now, in a random draw of 1 ball, the probability that the ball drawn is black is $$\frac{\text{number of ways in which the favourable event occurs}}{\text{total number of outcomes}}$$ where the event is the drawing of a black ball, which comes out to be $\frac{10}{30}$. So far, so good.

Now, suppose that we have another box, again with a small hole. Suppose also that 10 identical black balls and 20 identical white balls are put into the box. What is the probability now, that a randomly drawn ball will be black?

According to my understanding, (which I'm not too sure about) there are only 2 distinct outcomes - drawing a white ball and drawing a black ball, because all the balls are identical. From that point of view the probability should be $\frac{1}{2}$.

On the other hand, it just doesn't seem right. This logic says that whatever be the number of items of each type, only the number of types can affect the probability. But that doesn't seem realistic. If I hade a huge number of white balls, and only 1 black ball the probability of getting a black ball would be very very small. But this result says otherwise.

I'm in a dilemma here, can any one sort out this problem?

Aritra Das
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1 Answers1

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The answer is again $\frac{1}{3}$, as can be seen from the fact that in your answer to the first part you did not [need to] rely on the distinctness of the white (black) balls from others of the same colour.

Suppose some outside agent has labelled the balls from $1,\dots,30$ in some way that is undetectable to the touch. Now the balls are again distinct, and the probability of drawing a black ball is again $\frac{10}{30}$. The mere act of labeling the balls did not change their probability of selection (since it was undetectable to you), so this probability must be the same as for unlabeled balls.

Probability of choosing 1 black ball and 0 white balls from 10 non-identical black balls and 20 non-identical white balls is:

$\Pr[X=1] = \dfrac{{10 \choose 1}{20 \choose 0}}{{30 \choose 1}} = \dfrac{10 \times 1}{30} = \dfrac{10}{30}$

If on the other hand the black balls are identical and we wish to rely on this fact, the answer is obtained by simple ratio of number of black balls to the total:

$\Pr[X=1] = \dfrac{10}{30}$

Marconius
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  • "you did not rely on the distinctness ... " I did actually. Because I knew they were distinct, I knew that selecting black ball B1 was distinct from selecting black ball B2, or B3, or ... And there were 10 such selections, similarly 20 for white giving the probability of $\frac{10}{20}$ – Aritra Das Aug 04 '15 at 16:36
  • But that would mean that identical objects and distinct objects are same, but that is not the case. What I mean by that is there are situations where, if identical objects are assumed to be distinct, the probability calculated would be incorrect. For example suppose there are n identical objects. You have to select any 2 of them. It can be done in only 1 way, whereas if the objects were distinct, the selection could be done in $^nC_2$ ways. – Aritra Das Aug 04 '15 at 18:06
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    Okay, when counting number of ways the distinction between distinct and non-identical may be important. When calculating probabilities, there are often many equivalent approaches (see answer). This problem does not warrant a more complicated approach such as stars and bars, e.g. see http://math.stackexchange.com/questions/290420/selecting-5-doughnuts-counting-problem – Marconius Aug 04 '15 at 19:03
  • Okay, I understand what you mean. Just one last thing, what if instead of 1, 2 balls were to be selected? The probability required is that of getting 2 black balls. Would then too the identical nature or distinct nature not matter? – Aritra Das Aug 04 '15 at 19:43
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    You can compute $\dfrac{{10 \choose 2}{20 \choose 0}}{{30 \choose 2}} = \dfrac{10 \times 9 / 2}{30 \times 29 / 2} = \dfrac{10\times9}{30\times29}$ if considering non-identical balls, and $\dfrac{10}{30} \times \dfrac{9}{29}$ otherwise. In both cases, selections are made without replacement. Consider the black and white balls as equivalence classes - it is not important to the problem whether they can be differentiated from others of their type or not. – Marconius Aug 04 '15 at 19:56