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# PS:m23#22

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Manager
Joined: 05 Jan 2009
Posts: 76
PS:m23#22 [#permalink]

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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: 103
Location: Brazil
Re: PS:m23#22 [#permalink]

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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: 144
Re: PS:m23#22 [#permalink]

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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: 188
Schools: Stanford...Wait, I will come!!!
Re: PS:m23#22 [#permalink]

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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: 103
Location: Brazil
Re: PS:m23#22 [#permalink]

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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...

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: PS:m23#22   [#permalink] 22 Jul 2009, 05:49
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# PS:m23#22

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