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Re M2322 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M2322 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M2322 [#permalink]
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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 [#permalink]
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30 Jun 2018, 12:55
When you say 2 different numbers are randomly selected from set means it cannot be {1,1}, {2,2}, {3,3}, {4,4}, and {5,5}. But it can be {1,2} or {2,1}.
The question should clearly indicate that subsets have to be different.
Thanks.










