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Math Expert
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If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4? A. \(\frac{3}{5}\) B. \(\frac{7}{10}\) C. \(\frac{4}{5}\) D. \(\frac{9}{10}\) E. \(\frac{19}{20}\)
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Re M2322
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16 Sep 2014, 01:19
Official Solution:If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4? A. \(\frac{3}{5}\) B. \(\frac{7}{10}\) C. \(\frac{4}{5}\) D. \(\frac{9}{10}\) E. \(\frac{19}{20}\) Let's find the probability of the opposite event and subtract that value from 1. The opposite event would be if we choose 2 different number so that their sum will be less or equal to 4. The number of total outcomes is \(C^2_5=10\) (choosing 2 different numbers from the set of 5 different numbers); The number of outcomes when \(sum\leq{4}\) is 2: only (1,2) and (1,3); \(P=1\frac{2}{10}=\frac{4}{5}\). Answer: C
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Re: M2322
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11 Jun 2016, 10:52
Bunuel wrote: Official Solution:
If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4?
A. \(\frac{3}{5}\) B. \(\frac{7}{10}\) C. \(\frac{4}{5}\) D. \(\frac{9}{10}\) E. \(\frac{19}{20}\)
Let's find the probability of the opposite event and subtract that value from 1. The opposite event would be if we choose 2 different number so that their sum will be less or equal to 4. The number of total outcomes is \(C^2_5=10\) (choosing 2 different numbers from the set of 5 different numbers); The number of outcomes when \(sum\leq{4}\) is 2: only (1,2) and (1,3); \(P=1\frac{2}{10}=\frac{4}{5}\).
Answer: C Thank you for this great questions, gmat club is really good! This said, why do I have only 2 possible outcomes? (1,2) and (1,3).. why doesn't (2,1) and (3,1) work? I can randomly select them in this order as well?



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Re: M2322
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11 Aug 2016, 04:21
By definition a subset contains only different elements of the subset. So {1,2} and {2,1} is the same subset.



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Re: M2322
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04 Jul 2017, 08:35
[/m] Bunuel wrote: Official Solution:
If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4?
A. \(\frac{3}{5}\) B. \(\frac{7}{10}\) C. \(\frac{4}{5}\) D. \(\frac{9}{10}\) E. \(\frac{19}{20}\)
Let's find the probability of the opposite event and subtract that value from 1. The opposite event would be if we choose 2 different number so that their sum will be less or equal to 4. The number of total outcomes is \(C^2_5=10\) (choosing 2 different numbers from the set of 5 different numbers); The number of outcomes when \(sum\leq{4}\) is 2: only (1,2) and (1,3); \(P=1\frac{2}{10}=\frac{4}{5}\).
Answer: C The answer is wrong \(P=1\frac{2}{10}=\frac{8}{5}\). and not\(\frac{4}{5}\)



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Re: M2322
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04 Jul 2017, 08:37
pranab223 wrote: [/m] Bunuel wrote: Official Solution:
If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4?
A. \(\frac{3}{5}\) B. \(\frac{7}{10}\) C. \(\frac{4}{5}\) D. \(\frac{9}{10}\) E. \(\frac{19}{20}\)
Let's find the probability of the opposite event and subtract that value from 1. The opposite event would be if we choose 2 different number so that their sum will be less or equal to 4. The number of total outcomes is \(C^2_5=10\) (choosing 2 different numbers from the set of 5 different numbers); The number of outcomes when \(sum\leq{4}\) is 2: only (1,2) and (1,3); \(P=1\frac{2}{10}=\frac{4}{5}\).
Answer: C The answer is wrong \(P=1\frac{2}{10}=\frac{8}{5}\). and not\(\frac{4}{5}\) Really? Check again, please.
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Re: M2322
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04 Jul 2017, 09:07
Bunuel wrote: pranab223 wrote: [/m] Bunuel wrote: Official Solution:
If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4?
A. \(\frac{3}{5}\) B. \(\frac{7}{10}\) C. \(\frac{4}{5}\) D. \(\frac{9}{10}\) E. \(\frac{19}{20}\)
Let's find the probability of the opposite event and subtract that value from 1. The opposite event would be if we choose 2 different number so that their sum will be less or equal to 4. The number of total outcomes is \(C^2_5=10\) (choosing 2 different numbers from the set of 5 different numbers); The number of outcomes when \(sum\leq{4}\) is 2: only (1,2) and (1,3); \(P=1\frac{2}{10}=\frac{4}{5}\).
Answer: C The answer is wrong \(P=1\frac{2}{10}=\frac{8}{5}\). and not\(\frac{4}{5}\) Really? Check again, please. You are right its \(\frac{4}{5}\)



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Re: M2322
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16 Jul 2017, 00:42
How is it not B =7/10?
When you consider the case , how many pairs of numbers whose sum is greater than 4, you get
(1,4),(1,5).(2,4),(2,5),(2,3),(3,4),(3,5),(4,5) which is 7 outcomes out of total possible 10 outcomes.
Please explain.



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Re: M2322
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16 Jul 2017, 04:12
ajanpaul wrote: How is it not B =7/10?
When you consider the case , how many pairs of numbers whose sum is greater than 4, you get
(1,4),(1,5).(2,4),(2,5),(2,3),(3,4),(3,5),(4,5) which is 7 outcomes out of total possible 10 outcomes.
Please explain. It's not that hard to do the following: 1  (1,4) 2  (1,5) 3  (2,4) 4  (2,5) 5  (2,3) 6  (3,4) 7  (3,5) 8  (4,5)
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Re: M2322
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15 Mar 2018, 09:25
Only (1,2) and (1,3) are less than or equal to 4. The required probability is 1(2/5C2). The answer is 4/5.
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Re: M2322
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20 Jun 2019, 12:28
Bunuel wrote: Official Solution:
If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4?
A. \(\frac{3}{5}\) B. \(\frac{7}{10}\) C. \(\frac{4}{5}\) D. \(\frac{9}{10}\) E. \(\frac{19}{20}\)
Let's find the probability of the opposite event and subtract that value from 1. The opposite event would be if we choose 2 different number so that their sum will be less or equal to 4. The number of total outcomes is \(C^2_5=10\) (choosing 2 different numbers from the set of 5 different numbers); The number of outcomes when \(sum\leq{4}\) is 2: only (1,2) and (1,3); \(P=1\frac{2}{10}=\frac{4}{5}\).
Answer: C Bunuel can you please explain why the order does not matter?



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Re: M2322
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20 Jun 2019, 14:47
A quick way to test whether or not order matters is to try making the same selection in two different orders to see if you get a new result.
For example, if I am making a 3person group from A, B, C, D, and E and I select A, then B, then C, I get the same group as when I select C, then B, then A. Same result = order doesn't matter = combination.
On the other hand, if I am selecting 1st, 2nd, and 3rd place from A, B, C, D, and E, and I select A, then B, then C, I get a different order than when I select C, then B, then A. Different result = order does matter = permutation.
Trying that here, if I am picking 2 numbers from {1,2,3,4,5}, and I pick 1, then 2, I get the same sum (3) as when I pick 2, then 1. Same result = order doesn't matter = combination.
While there are a few key words that indicate whether or not order matters, we don't always get them, so the best way to figure it out is to test it yourself.



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Re: M2322
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04 Jul 2019, 05:19
the answer calculated is based on assumption greater than or equal to 4 not only greater than 4.
If only greater than 4 the answer is 7/10 for sure.
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Re: M2322
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04 Jul 2019, 05:30
anand.prakash90@gmail.com wrote: the answer calculated is based on assumption greater than or equal to 4 not only greater than 4.
If only greater than 4 the answer is 7/10 for sure.
Thanks No, the answer is correct as it is. We can get the sum less than or equal to 4 in only two cases (1,2) and (1,3), the probability of that is 2/10. All other cases give the sum strictly MORE than 4, so the probability of more than 4 is 1  2/10 = 4/5.
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Re: M2322
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04 Jul 2019, 07:56
Bunuel wrote: If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4?
A. \(\frac{3}{5}\) B. \(\frac{7}{10}\) C. \(\frac{4}{5}\) D. \(\frac{9}{10}\) E. \(\frac{19}{20}\) total pairs ; 5c2 ; 10 and <=4 ; (1,2) & ( 1,3) so pairs; 2/10 ; 1/5 and pairs >4 11/5; 4/5 IMO C
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