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# A black sack contains three green balls, five yellow balls,

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SVP
Joined: 03 Feb 2003
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A black sack contains three green balls, five yellow balls, [#permalink]  03 Jul 2003, 12:30
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A black sack contains three green balls, five yellow balls, and four white ones. Three balls are taken at random without repetition. What is the probability of having all the the three balls of different colors?

An easy stuff for night rumination...
Intern
Joined: 23 Jun 2003
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Re: PS: PROBA-6 [#permalink]  03 Jul 2003, 19:55
Here's what I got:
3/12 * 5/11 * 4/10 = 1/22

stolyar wrote:
A black sack contains three green balls, five yellow balls, and four white ones. Three balls are taken at random without repetition. What is the probability of having all the the three balls of different colors?

An easy stuff for night rumination...
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 78 [0], given: 0

This is an example of so-called classical hypergeometrical distribution:

3C1*5C1*4C1/12C3 = 3/11
Manager
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Stolyar, is this approach fair?:

3 G, 5Y, 4W

P(green) = 3/12
and
p(yellow) = 5/11
and
p(white) = 4/10

(3/12)x(5/11)x(4/10) = 1/22

Then because the 3 colors could be drawn out 3! ways:

3! x (1/22) = 6/22 = 3/11
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 78 [0], given: 0

JP wrote:
Stolyar, is this approach fair?:

3 G, 5Y, 4W

P(green) = 3/12
and
p(yellow) = 5/11
and
p(white) = 4/10

(3/12)x(5/11)x(4/10) = 1/22

Then because the 3 colors could be drawn out 3! ways:

3! x (1/22) = 6/22 = 3/11

your approach is legal, but mine is quicker (I think so)
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