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If two different numbers are to be selected from set {1, 2,
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29 Jul 2008, 10:37
. If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6}, what is the probability that the sum of two numbers is a perfect square number? A. 1/2 B. 1/3 C. 1/9 D. 2/9 E. 4/9 here what confuses me is , in the total poss cases, do we take ,for eg 1,3 and also along with that 3,1 or should they be considered only once? == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



Intern
Joined: 26 May 2008
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Re: probability
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29 Jul 2008, 10:51
Here perfect squares can be either 4 or 9.
4 = 3+1 or 1+3 (2 ways) 9 = 3+6 or 6 +3 (2 ways) = total 4 ways
Total ways of selecting to numbers = 6 *6 = 36 ways
Required probabilty = 4/36 = 1/9



Current Student
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Re: probability
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29 Jul 2008, 10:57
what abot 4,5 and 5,4? also total ways of selecting two nos. from 6 is 6c2 rite?



Senior Manager
Joined: 14 Mar 2007
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Re: probability
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29 Jul 2008, 11:02
We have 15 possible option and 5 good option. Probability=5/15=1/3



Current Student
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Re: probability
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29 Jul 2008, 11:10
how 5 good option? it shud be 6 rite?



Senior Manager
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Re: probability
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29 Jul 2008, 12:09
I think we have 5 good options: (1,3) (2,6) (3,5) (3,6) (4,5)
I hpe these are the perfect squere numbers: 4, 8, 9, ...
I am not sure, may be I am not correct.



GMAT Tutor
Joined: 24 Jun 2008
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Re: probability
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29 Jul 2008, 17:39
arjtryarjtry wrote: . If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6}, what is the probability that the sum of two numbers is a perfect square number? A. 1/2 B. 1/3 C. 1/9 D. 2/9 E. 4/9
here what confuses me is , in the total poss cases, do we take ,for eg 1,3 and also along with that 3,1 or should they be considered only once? 2^2 = 4 and 3^2 = 9 are the only perfect squares we can get as a sum of two numbers from the set. Note that 8 is not a perfect square (8 is a cube, however). Now, we can either assume order matters, or assume it doesn't we just need to be consistent. If order matters, there are 6*5 = 30 different pairs of numbers we can choose. Again, if order matters, we can get squares in the following ways: 1,3 3,1 3,6 4,5 5,4 6,3 So the probability should be 6/30 = 1/5. That isn't among the answer choices where is the question from?
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Intern
Joined: 01 Jul 2008
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Re: probability
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30 Jul 2008, 00:14
number of ways that you can select 2 numbers out of 6 would be 6C2=15 ways favourable outcomes are {1,3}, {4,5}, & {3,6} = 3 ways
hence the probability would be 3/15=1/5 (well it is not one of the options!!!)



Senior Manager
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Re: probability
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30 Jul 2008, 00:18
hi0parag,
Thank you for correcting my mistake.



Manager
Joined: 27 May 2008
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Re: probability
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30 Jul 2008, 03:26
Wrong Question. The answer is 1/5 == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.




Re: probability &nbs
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30 Jul 2008, 03:26






