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I got this question in an exam a while ago and I wasn't sure how exactly I could solve it:

An urn contains 3 red balls, 3 blue balls and 3 green balls. Three are drawn at 
random from the urn. What's the probability they're all of a different colour?

It is not specified if with or without replacement. Could someone please clarify me on both variants? Thanks :)

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If they just say "three are drawn", I'd assume they mean without replacement. – Raskolnikov Aug 23 '12 at 12:32
up vote 4 down vote accepted

If without replacement, 1st ball could be any, 2nd could be any of 6 of the 8, 3rd any of 3 of the 7, so $(6/8)(3/7)=9/28$.

If with replacement, 1st could be any, 2nd any of 6 of the 9, 3rd any of 3 of the 9, so $(6/9)(3/9)=2/9$.

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