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PS:m23#22

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Manager
Joined: 05 Jan 2009
Posts: 82
Followers: 1

Kudos [?]: 116 [0], given: 2

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18 Jul 2009, 14:17
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please discuss the PS below.
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Manager
Joined: 03 Jul 2009
Posts: 106
Location: Brazil
Followers: 3

Kudos [?]: 87 [0], given: 13

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18 Jul 2009, 16:28
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 $$8/10 = 4/5$$.
Manager
Joined: 27 Jun 2008
Posts: 158
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Kudos [?]: 30 [0], given: 11

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22 Jul 2009, 00:55
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
Manager
Joined: 14 Nov 2008
Posts: 195
Schools: Stanford...Wait, I will come!!!
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Kudos [?]: 105 [0], given: 3

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22 Jul 2009, 02:35
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
Manager
Joined: 03 Jul 2009
Posts: 106
Location: Brazil
Followers: 3

Kudos [?]: 87 [0], given: 13

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22 Jul 2009, 05:49
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, 05:49
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PS:m23#22

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