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Re: In a certain box there're 3 red, 5 blue, and 2 green balls. [#permalink]
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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Re: In a certain box there're 3 red, 5 blue, and 2 green balls. [#permalink]
The above is a classical example of so-called hypergeometrical distribution.
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Re: In a certain box there're 3 red, 5 blue, and 2 green balls. [#permalink]
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gravedigger wrote:
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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