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m10 #14 another solution

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Senior Manager
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Joined: 18 Aug 2009
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m10 #14 another solution [#permalink]

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New post 15 Nov 2009, 02:55
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Found the OE bit complicated than usual probability q approach... so posting another solution which maybe easier to understand for some (who fall in similar category:) )

Set \(S\) consists of all prime integers less than 10. If a number is selected from set \(S\) at random and then another number, not necessarily different, is selected from set \(S\) at random, what is the probability that the sum of these numbers is odd?

* \(\frac{1}{8}\)
* \(\frac{1}{6}\)
* \(\frac{3}{8}\)
* \(\frac{1}{2}\)
* \(\frac{5}{8}\)

Total ways to choose 2 numbers (repeat doesn't matter): 4x4=16
This would have been 4x3=12 had the Q not mentioned "not necessarily different". I missed that part in the test :(
Result is odd only when one number is 2 and the other any of the others.

Favorable cases when "2" is the first to be drawn {(2,3),(2,5),(2,7)}=3 , another 3 when "2" is the second to be drawn.

So, probability=(favorable cases/total cases)=6/16=3/8

Kudos [?]: 354 [1], given: 9

Manager
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Joined: 24 Jul 2009
Posts: 73

Kudos [?]: 147 [0], given: 124

Location: United States
GMAT 1: 590 Q48 V24
Re: m10 #14 another solution [#permalink]

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New post 15 Nov 2009, 03:14
Simple Explanation, Kudos !!
I hate Probability :evil:

Kudos [?]: 147 [0], given: 124

Senior Manager
Senior Manager
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Joined: 18 Aug 2009
Posts: 299

Kudos [?]: 354 [0], given: 9

Re: m10 #14 another solution [#permalink]

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New post 15 Nov 2009, 06:14
ctrlaltdel wrote:
Simple Explanation, Kudos !!
I hate Probability :evil:


Thanks :-D
don't like it either, but something unavoidable!

Kudos [?]: 354 [0], given: 9

Re: m10 #14 another solution   [#permalink] 15 Nov 2009, 06:14
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