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Legato
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21 Jun 2010, 9:53 pm

Hey, I've been struggling with something literally all day, and I believe the key lies in the answer to this question: Given a six-sided die and three attempts, what is the probability of getting a 2 or 1 on at least two tries?

An even simpler example in case I'm not illustrating this well enough: Given a purely random selection of three colors (Red, Green, and Blue), what is the probability of getting Red at least twice out of three tries?

Can you please provide the formula you used to calculate this? Thanks a trillion.



sgrannel
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21 Jun 2010, 11:30 pm

you could get all three, which is 1/27. You could get two out of three three different ways, each with a probability of 1/3x/1/3x2/3 = 2/27. The total probability is 7/27.



ari_
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21 Jun 2010, 11:47 pm

sgrannel wrote:
you could get all three, which is 1/27. You could get two out of three three different ways, each with a probability of 1/3x/1/3x2/3 = 2/27. The total probability is 7/27.


He's right, the probability of that happening is 7/27.