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# There are 3 green; 4 yellow; and 5 blue balls in a sack. 6

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Director
Joined: 12 Jun 2006
Posts: 531

Kudos [?]: 166 [0], given: 1

There are 3 green; 4 yellow; and 5 blue balls in a sack. 6 [#permalink]

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06 Feb 2007, 23:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are 3 green; 4 yellow; and 5 blue balls in a sack. 6 balls are taken at random w/o replacement. Find the prob to have 1 green; 2 yellow; and 3 blue balls.

Kudos [?]: 166 [0], given: 1

Director
Joined: 06 Feb 2006
Posts: 893

Kudos [?]: 129 [0], given: 0

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07 Feb 2007, 01:15
God this is time consuming question...

Kudos [?]: 129 [0], given: 0

Manager
Joined: 10 Dec 2005
Posts: 112

Kudos [?]: 5 [0], given: 0

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07 Feb 2007, 15:45
ggarr wrote:
There are 3 green; 4 yellow; and 5 blue balls in a sack. 6 balls are taken at random w/o replacement. Find the prob to have 1 green; 2 yellow; and 3 blue balls.

The answer would be I think

3 * 6 * 10/12 * 11 * 9

5/33
_________________

"Live as if you were to die tomorrow. Learn as if you were to live forever." - Mahatma Gandhi

Kudos [?]: 5 [0], given: 0

Senior Manager
Joined: 04 Jan 2006
Posts: 276

Kudos [?]: 44 [0], given: 0

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07 Feb 2007, 17:07
ggarr wrote:
There are 3 green; 4 yellow; and 5 blue balls in a sack. 6 balls are taken at random w/o replacement. Find the prob to have 1 green; 2 yellow; and 3 blue balls.

Let A = Having 1 green, 2 yellow, and 3 blue balls

Prob(A) = (Total ways to pick 1 Green, 2 Yellow, and 3 Blue)/(Total ways to pick 6 balls randomly)

Total ways to pick 6 balls randomly = 12C6
Total ways to pick 1 G, 2 Y, and 3B = (3C1) x (4C2) x (5C3)

Prob(A) = ((3C1) x (4C2) x (5C3))/12C6
= (3) x (2x3)x (5x2)/ (11 x 2 x 3 x 2 x 7)
= (3 x 5)/(11 x 7) = 15/77

15/77 is the answer for me.

Kudos [?]: 44 [0], given: 0

Re: Prob prob II   [#permalink] 07 Feb 2007, 17:07
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# There are 3 green; 4 yellow; and 5 blue balls in a sack. 6

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