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I have $\ 30 $ marbles. $\ 25 $ are white, $\ 3 $ are blue and $\ 2 $ are red. same color marbles are identical.

If I pick randomly and without replacement $\ 4 $ marbles, what is the probability that I'll pick two each two of two colors?

Trying to make it easier, I assumed all marbles are different, so there are $\ 30 \cdot 29 \cdot 28 \cdot 27 $ ways to pick them and then number of options for :

Picking $2$ blue and $2$ red marbles are $\ {3 \choose 2}{25 \choose 2} \cdot 4! $ options.

Picking $2$ blue and $2$ white marbles are $\ {3 \choose 2}{2 \choose 2} \cdot 4! $ options.

Picking $2$ white and $2$ red marbles are $\ {25 \choose 2}{2 \choose 2 }\cdot 4! $.

The three events are mutually exclusive, so I should be able to just add them all together but that's the wrong answer. Any suggestions?

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    $\begingroup$ I assume 2 white and 2 white isn't an option? $\endgroup$ – Arthur Oct 31 '18 at 12:39
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    $\begingroup$ Do you mean two each of two colors? $\endgroup$ – N. F. Taussig Oct 31 '18 at 12:40
  • $\begingroup$ Correct. Sorry I try to be clear as I could, I translated this question to english. 4 marbles, 2 different colors, 2 marlbes of each color $\endgroup$ – bm1125 Oct 31 '18 at 12:40
  • $\begingroup$ So what is the final answer you get, and what is the answer you should've gotten? $\endgroup$ – Arthur Oct 31 '18 at 13:13
  • $\begingroup$ Calculating again, I suddenly got different answer on my calculator which fit one of the possible answers ( $\ \frac{401}{9135} $ ). But I'm not sure though if it is the answer? $\endgroup$ – bm1125 Oct 31 '18 at 13:19
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Correct answer is $\frac{401}{9135}$. It is calculated as follows:$\frac{\binom{25}{2}*\binom{3}{2}}{\binom{30}{4}}+\frac{\binom{25}{2}*\binom{2}{2}}{\binom{30}{4}}+\frac{\binom{3}{2}*\binom{2}{2}}{\binom{30}{4}}$

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You are drawing a random $4$-element subset from a set $S$ containing $30$ elements. (Even though some elements of $S$ look alike the elements of $S$ are "secretly" numbered: $1$$25$ for the white ones, $26$$28$ for the blue ones, and $29$$30$ for the red ones.) There are three kinds of "good" subsets. You have counted them, but for no reason multiplied the numbers by $4!$. (Note that the order in which the four marbles are drawn plays no rôle. All four are put in a smaller bag.) The probablity you are after is the total number of "good" subsets divided by the total number of all $4$-element subsets.

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  • $\begingroup$ I multiplied by $\ 4! $ because as you said I have the elements of S are "secretly" numbered and therefore I want to count both $\ R,R,B,B $ and $\ B,B,R,R $ . I mean maybe if I had chosen instead of $\ 30 \cdot 29 \cdot 28 \cdot 27 $ a $\ 30!/(26!4!) $ I wouldn't have to multiply by $\ 4! $ any event? $\endgroup$ – bm1125 Oct 31 '18 at 13:37

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