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In a box there are 10 white ball and 10 black balls.Two balls are successively drawn.What is the probability to get two balls of different colors?

My interpretation is that the order doesn't matter, because the important is that the balls have different colors.So to find the possible cases I computed a combination, $^{20}C_{2}$. To find the favorable cases I multiplied $10$ white balls by $10$ black balls.

The final mathematical expression stays: $\frac{10^2 \cdot 2}{20 \cdot 19}$

The book's interpretation is the following:

For the possible cases, there are $20$ for the first ball and $19$ for the second.So there are $20 \cdot 19$ possible cases.Regarding the favorable cases the book says that there are $10 \cdot 10 \cdot 2$.

It's clear that by the book's interpretation the order matters.The two final mathematical expressions are the same.How can this happen if the interpretations were different?

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Because the final answer (i.e. the probability), should not depend on how you solve the problem. You just listed two different ways to solve the same problem, so the answers coincide. –  Tom Artiom Fiodorov Feb 26 '12 at 19:33

1 Answer 1

up vote 2 down vote accepted

The basis of this calculation is that there are equally probable patterns, and taking the ratio of those patterns meeting the condition to the total number gives the probability of the condition. So let us take the ratio of numbers of patterns (where order matters as does identification of the different balls).

We can start by looking at all patterns from arranging all twenty balls. There are $20!$ ways of ordering all twenty balls. Of these $10 \times 10 \times 18!$ have a black followed by a white at the beginning and $10 \times 10 \times 18!$ have a white followed by a black at the beginning, so the proportion of patterns meeting the condition is $\dfrac{10 \times 10 \times 18!+10 \times 10 \times 18!}{20!}$.

The book then decides that all the arrangements of the last eighteen can be treated as equivalent and grouped together as equivalence classes, with each equivalence class having $18!$ elements, to give a proportion of equivalence classes of $\dfrac{10 \times 10 +10 \times 10 }{20 \times 19}$.

You instead decide that all arrangements of the last eighteen and all arrangements of the first two can be treated as equivalent and grouped together as equivalence classes, with each equivalence class having $18!\times 2!$ elements, to give a proportion of equivalence classes of $\dfrac{10 \times 10 }{20 \times 19/2!}$.

The only difference between you and the book is that you have half as many equivalence classes and they are each twice the size.

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