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PS:m23#22

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Manager
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Joined: 05 Jan 2009
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Kudos [?]: 11 [0], given: 2

PS:m23#22 [#permalink] New post 18 Jul 2009, 13:17
00:00
A
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Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
please discuss the PS below.
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Manager
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Joined: 03 Jul 2009
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Kudos [?]: 49 [0], given: 13

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Re: PS:m23#22 [#permalink] New post 18 Jul 2009, 15:28
From 2C5 we know that we have 10 possibilities.

_ _ => the two numbers

5 _ => 4 possibilities of the sum is more than 4
4 _ => 3 possibilities of the sum is more than 4
3 _ => 1 possibilities of the sum is more than 4

We do not need to do with 2 and 1 because the possibilities will be repeated. Thus we have 8 possibilities in a total of 10. This is 8/10 = 4/5.
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Re: PS:m23#22 [#permalink] New post 21 Jul 2009, 23:55
another option would be to look at the combination which can have total less than or equal to 4
it will be {1,2} and {1,3} out of total 10 combination {5C2}. THe probability would be
1-2/10 = 4/5
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Re: PS:m23#22 [#permalink] New post 22 Jul 2009, 01:35
pmal04 wrote:
please discuss the PS below.

the sample space is 5C2
now, if the number of way, the sum will be less than or equal to 4, is 1,2
1,3.
so, 8/10=4/5
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Re: PS:m23#22 [#permalink] New post 22 Jul 2009, 04:49
Yes, the option suggested by irajeevsingh is a little bit faster. And in the GMAT you will like those seconds...
Re: PS:m23#22   [#permalink] 22 Jul 2009, 04:49
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PS:m23#22

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