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5 blue marbles, 3 red marbles and 4 purple marbles are

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5 blue marbles, 3 red marbles and 4 purple marbles are [#permalink] New post 25 Jan 2008, 09:44
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5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn without replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?

A) 4/33
B) (5/36)^2
C) 1/2
D) (31/36)^2
E) 29/33
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Re: PS: Probability [#permalink] New post 25 Jan 2008, 12:11
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A

\(p=\frac{C^5_2*C^4_2*P^4_4}{P^{12}_4}=\frac{5!*4!*4!*8!}{3!*2!*2!*2!*12!}=\frac{5*4*4*3*4*3}{2*12*11*10*9}=\frac{4}{33}\)
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Re: PS: Probability [#permalink] New post 25 Jan 2008, 12:42
netcaesar,

where did you find this problem ?
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Re: PS: Probability [#permalink] New post 25 Jan 2008, 18:14
walker wrote:
A

\(p=\frac{C^5_2*C^4_2*P^4_4}{P^{12}_4}=\frac{5!*4!*4!*8!}{3!*2!*2!*2!*12!}=\frac{5*4*4*3*4*3}{2*12*11*10*9}=\frac{4}{33}\)



well i guess u mean to say the answer is :1 - 4/33 = 29/33
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Re: PS: Probability [#permalink] New post 25 Jan 2008, 21:39
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You are right. It's very important carefully to read question....
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Re: PS: Probability [#permalink] New post 26 Jan 2008, 08:40
I found this problem in the GMATPrep, (http://www.mba.com)
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Re: PS: Probability [#permalink] New post 26 Jan 2008, 20:24
netcaesar wrote:
5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn without replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?

A) 4/33
B) (5/36)^2
C) 1/2
D) (31/36)^2
E) 29/33


Probability that there will be 2 blue and 2 purple.

B B P P

5/12 * 4/11 * 4/10*3/9 ---> 2/99

Now we have 4!/2!2! ways to arrange BBPP so thats 6

6*2/99 --> 4/33 Now its just 1-4/33 = 29/33

E
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Re: PS: Probability [#permalink] New post 26 Jan 2008, 23:34
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netcaesar wrote:
What means P12, 4???


\(P^{12}_4\) - the total number of permutations 4 of 12=5+3+4
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Re: PS: Probability   [#permalink] 26 Jan 2008, 23:34
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