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3 red balls, 4 green balls, and 5 yellow balls are in a bag

3 red balls, 4 green balls, and 5 yellow balls are in a bag. If 2 balls are drawn randomly (without replacement) find the probability that the balls are different colors.

I got 47/66 that the balls are different colors...is that right?
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Answer #1
http://www.futureaccountant.com/probability/problems-solutions/drawing-picking-selecting-two-three-more-balls-box-bag-urn-04.php and many variations of the same bag.
answered by: tareeq
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Answer #2
Thank you very much for you help:)
answered by: Sasuke
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Answer #3
Correct! Well-done!

P(RR)=(3/12)(2/11)=6/132
P(GG)=(4/12)(3/11)=12/132
P(YY)=(5/12)(4/11)=20/132
Probability to draw two of same colour
=(6+12+20)/132=19/66
=>
Probability to draw two of different colours
=(66-19)/66=47/66
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