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October 31, 2014

October 31, 2014

Posted by **Ryan** on Sunday, November 29, 2009 at 7:59pm.

- Probability again -
**economyst**, Monday, November 30, 2009 at 9:07amfollow the basic methodology that Drwls showed you in the previous post. Assume each color ball can be distinguished from each other. (e.g., name the green balls g1 and g2) First, the denominator. How many different ways can 3 balls from 12 be chosen. 12-choose-3 is 12!/3!(12-3)!

Now the numerator. How many different ways can 3 red balls be chosen from 6? Plus how many different ways can 3 blue balls be chosen from 4.

Take it from here.

- Probability again -
**Ryan**, Monday, November 30, 2009 at 7:24pmim lost

- Probability again -
**ali**, Monday, July 30, 2012 at 2:52pmA bag contains 5 yellow balls, 5 orange balls, 6 white balls and 6 red balls. Five balls are drawn and NOT replaced each time

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