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tino
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In a certain box there're 3 red, 5 blue, and 2 green balls. 29 Apr 2003, 14:04
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In a certain box there're 3 red, 5 blue, and 2 green balls. 6 balls are taken randomly from the box. What is the probability that the taken balls will contain 2 red, 2 blue, and 2 green ones?



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In a certain box there're 3 red, 5 blue, and 2 green balls. 30 Apr 2003, 06:58
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Ans is (3C2)(5C2)(2c2)/10C6.

Ans is 1/7
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In a certain box there're 3 red, 5 blue, and 2 green balls. 30 Apr 2003, 13:40
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:) Great, i too have 1/7
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In a certain box there're 3 red, 5 blue, and 2 green balls. 01 May 2003, 13:42
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Answer: 1/7

We have a total of 3+5+2 = 10 balls. If we take 6 balls, we have 10C6 total combinations, which equals 210.

Now, the fav combinations are best found the following way: we take each color separately and see how many ways there're to fill the given color's quota with different balls:

Red: 3C2 = 3
Blue: 5C2 = 10
Green: 2C2 = 1

Each combination within a given color we can mix with any other combinations of the two remaining color to get the total combination. For example, if red is filled with balls Red#1 and Red#2, we can mix it with Blue#1 and Blue#2, Blue#1 and Blue#3, etc. So, the total number of fav combinations is a multiplication of those numbers:

Fav = 3 * 10 * 1 = 30
Prob = 30/210 = 1/7
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In a certain box there're 3 red, 5 blue, and 2 green balls. 04 May 2003, 23:04
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The above is a classical example of so-called hypergeometrical distribution.
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In a certain box there're 3 red, 5 blue, and 2 green balls. 05 Jun 2003, 16:09
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In a certain box there're 3 red, 5 blue, and 2 green balls. 6 balls are taken randomly from the box. What is the probability that the taken balls will contain 2 red, 2 blue, and 2 green ones?



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