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# If two different numbers are randomly selected from set { 1,

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Director
Joined: 07 Jun 2004
Posts: 605
Location: PA
If two different numbers are randomly selected from set { 1, [#permalink]

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Updated on: 13 Dec 2012, 02:07
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3
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Question Stats:

75% (00:47) correct 25% (03:54) wrong based on 22 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?

OA ia
4/5

My approach was

total outcomes 5 C 2 = 20

outcomes where sum > 4 are

1,4 1,5 = 2
2,3 2,4 2,5 = 3
3,2, 3,3 3,4 3,5 = 4

4 any of the 5 = 5
5 any of the 5 = 5

total 19

so P(>4) = 19/20

why is this wrong

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Originally posted by rxs0005 on 23 Jul 2010, 08:52.
Last edited by Bunuel on 13 Dec 2012, 02:07, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 06 Jul 2010
Posts: 88

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23 Jul 2010, 09:15
1
rxs005,
First problem is 5C2 = 10 .. Not 20..

Point 2, you sum should be greater than 4, so these are the following possibilities you have
({1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}) which leads to 8 pairs

So the probability is 8 on 10 or 4/5..
Director
Joined: 07 Jun 2004
Posts: 605
Location: PA

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23 Jul 2010, 09:46
you are right thanks!

the key is "different" numbers in the Q

i missed that
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Math Expert
Joined: 02 Sep 2009
Posts: 46296
If two different numbers are randomly selected from set { 1, [#permalink]

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23 Jul 2010, 13:06
rxs0005 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?

OA ia 4/5

My approach was

total outcomes 5 C 2 = 20

outcomes where sum > 4 are

1,4 1,5 = 2
2,3 2,4 2,5 = 3
3,2, 3,3 3,4 3,5 = 4

4 any of the 5 = 5
5 any of the 5 = 5

total 19

so P(>4) = 19/20

why is this wrong

Another approach: $$P(of \ event \ X)=1-P(of \ opposite \ event \ X)$$. Opposite event would be if we choose 2 different numbers so that their sum will be less or equal to 4.

# of total outcomes is $$C^2_5=10$$ (total # of choosing 2 different numbers from the set of 5 different numbers);

# of outcomes when $$sum\leq{4}$$ is (1,2) and (1,3), so 2.

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

Answer: $$\frac{4}{5}$$.

P. S. In the future pleas post the answer choices too.
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Intern
Joined: 18 Mar 2012
Posts: 13

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13 Dec 2012, 01:53
sridhar wrote:
rxs005,
First problem is 5C2 = 10 .. Not 20..

Point 2, you sum should be greater than 4, so these are the following possibilities you have
({1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}) which leads to 8 pairs

So the probability is 8 on 10 or 4/5..

How is the pare 1,4 GREATER than 4, I would say it is equal to four and therefore the answer should be 7/10.
Any argument ?
Intern
Joined: 18 Nov 2011
Posts: 35
Concentration: Strategy, Marketing
GMAT Date: 06-18-2013
GPA: 3.98

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10 Jan 2013, 21:54
arnijon90 wrote:
sridhar wrote:
rxs005,
First problem is 5C2 = 10 .. Not 20..

Point 2, you sum should be greater than 4, so these are the following possibilities you have
({1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}) which leads to 8 pairs

So the probability is 8 on 10 or 4/5..

How is the pare 1,4 GREATER than 4, I would say it is equal to four and therefore the answer should be 7/10.
Any argument ?

Because: 1+4 = 5 and 5>4
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Re: If two different numbers are randomly selected from set { 1, [#permalink]

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03 Dec 2017, 18:54
rxs0005 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?

The number of ways to select 2 numbers from 5 is 5C2 = (5 x 4)/2! = 20/2 = 10.

To find the number of sums that are greater than 4, we can find the number of sums that ARE NOT greater than 4 and subtract that from 10.

1, 2

1, 3

So there are 10 - 2 = 8 sums that are greater than 4.

So the probability that the sum of the two numbers is greater than 4 is 8/10 = 4/5.
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Joined: 27 Mar 2018
Posts: 11
Re: If two different numbers are randomly selected from set { 1, [#permalink]

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05 Apr 2018, 00:48
Bunuel wrote:
rxs0005 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?

OA ia 4/5

My approach was

total outcomes 5 C 2 = 20

outcomes where sum > 4 are

1,4 1,5 = 2
2,3 2,4 2,5 = 3
3,2, 3,3 3,4 3,5 = 4

4 any of the 5 = 5
5 any of the 5 = 5

total 19

so P(>4) = 19/20

why is this wrong

Another approach: $$P(of \ event \ X)=1-P(of \ opposite \ event \ X)$$. Opposite event would be if we choose 2 different numbers so that their sum will be less or equal to 4.

# of total outcomes is $$C^2_5=10$$ (total # of choosing 2 different numbers from the set of 5 different numbers);

# of outcomes when $$sum\leq{4}$$ is (1,2) and (1,3), so 2.

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

Answer: $$\frac{4}{5}$$.

P. S. In the future pleas post the answer choices too.

how we can solve this question? what is different between above question and this question.

What is the probability that the sum of two different numbers randomly picked (without replacement)
from the set S = {1, 2, 3, 4} is 5?
(A) 1/5
(B) 3/16
(C) 1/4
(D) 1/3
(E) 1/2
Math Expert
Joined: 02 Sep 2009
Posts: 46296
Re: If two different numbers are randomly selected from set { 1, [#permalink]

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05 Apr 2018, 00:56
VaibNop wrote:
Bunuel wrote:
rxs0005 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?

OA ia 4/5

My approach was

total outcomes 5 C 2 = 20

outcomes where sum > 4 are

1,4 1,5 = 2
2,3 2,4 2,5 = 3
3,2, 3,3 3,4 3,5 = 4

4 any of the 5 = 5
5 any of the 5 = 5

total 19

so P(>4) = 19/20

why is this wrong

Another approach: $$P(of \ event \ X)=1-P(of \ opposite \ event \ X)$$. Opposite event would be if we choose 2 different numbers so that their sum will be less or equal to 4.

# of total outcomes is $$C^2_5=10$$ (total # of choosing 2 different numbers from the set of 5 different numbers);

# of outcomes when $$sum\leq{4}$$ is (1,2) and (1,3), so 2.

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

Answer: $$\frac{4}{5}$$.

P. S. In the future pleas post the answer choices too.

how we can solve this question? what is different between above question and this question.

What is the probability that the sum of two different numbers randomly picked (without replacement)
from the set S = {1, 2, 3, 4} is 5?
(A) 1/5
(B) 3/16
(C) 1/4
(D) 1/3
(E) 1/2

The number of pairs of numbers from four numbers is 4C2 = 6:
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)

The number of pairs of numbers which add up to 5 is two:
(1, 4)
(2, 3)

The probability = fovorable/total = 2/6 = 1/3.

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Re: If two different numbers are randomly selected from set { 1,   [#permalink] 05 Apr 2018, 00:56
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