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Author Topic: help for a problem  (Read 7393 times)

ruiwca

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help for a problem
« on: October 22, 2010, 06:00:12 AM »
Hi all:

My child has a mathcount problem which I am not sure where to start:
Three fair six-sided number cubes, each with faces labeled 0,2,4,6,8,10, are rolled. what is the probabilit that the sum of the numbers rolled is greater than 20? Express your nswer as a common fraction.

Thanks.
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木牛流马

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Re: help for a problem
« Reply #1 on: October 22, 2010, 03:24:41 PM »
I got 35/216.
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Katalianna

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Re: help for a problem
« Reply #2 on: November 07, 2010, 11:23:38 AM »
I calculated and had 11/36 as my answer.
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t130152

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Re: help for a problem
« Reply #3 on: February 07, 2011, 01:55:55 PM »
The same thing as a+b+c>10 where 0<=a,b,c<=5 or x+y+z<=4 where 0<=x,y,z<=4.  It has the same counts as x+y+z+w=4 where 0<=x,y,z,w<=4, that is 7 chooses 3 = 35.
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wxc0212

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Re: help for a problem
« Reply #4 on: February 13, 2011, 10:50:33 PM »
Could somebody please explain in detail how they solved this question. How did you get 35/216 or 11/36
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