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# M23-22

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Math Expert
Joined: 02 Sep 2009
Posts: 58335

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16 Sep 2014, 01:19
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Difficulty:

45% (medium)

Question Stats:

61% (01:31) correct 39% (01:48) wrong based on 156 sessions

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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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Math Expert
Joined: 02 Sep 2009
Posts: 58335

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16 Sep 2014, 01:19
2
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);

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

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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);

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

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?

Intern
Joined: 22 Jul 2016
Posts: 1

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11 Aug 2016, 04:21
1
By definition a subset contains only different elements of the subset. So {1,2} and {2,1} is the same subset.
Intern
Joined: 14 Oct 2015
Posts: 6

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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);

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

$$P=1-\frac{2}{10}=\frac{8}{5}$$. and not$$\frac{4}{5}$$
Math Expert
Joined: 02 Sep 2009
Posts: 58335

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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);

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

$$P=1-\frac{2}{10}=\frac{8}{5}$$. and not$$\frac{4}{5}$$

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Intern
Joined: 14 Oct 2015
Posts: 6

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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);

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

$$P=1-\frac{2}{10}=\frac{8}{5}$$. and not$$\frac{4}{5}$$

You are right its $$\frac{4}{5}$$
Intern
Joined: 21 May 2017
Posts: 2

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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.

Math Expert
Joined: 02 Sep 2009
Posts: 58335

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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.

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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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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Intern
Joined: 27 Mar 2019
Posts: 38
Location: India
GMAT 1: 650 Q48 V32
GPA: 3

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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);

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

Bunuel can you please explain why the order does not matter?
Veritas Prep GMAT Instructor
Joined: 01 May 2019
Posts: 50

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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 3-person 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.
Intern
Joined: 21 Mar 2015
Posts: 8

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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.

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 58335

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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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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 1-1/5; 4/5
IMO C
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Re: M23-22   [#permalink] 04 Jul 2019, 07:56
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# M23-22

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