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

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

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New post Updated on: 13 Dec 2012, 02:07
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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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Re: Probability clarification [#permalink]

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New post 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..
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Re: Probability clarification [#permalink]

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

the key is "different" numbers in the Q

i missed that
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If two different numbers are randomly selected from set { 1, [#permalink]

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New post 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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Re: Probability clarification [#permalink]

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New post 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 ?
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Re: Probability clarification [#permalink]

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New post 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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New post 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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Re: If two different numbers are randomly selected from set { 1, [#permalink]

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New post 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
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Re: If two different numbers are randomly selected from set { 1, [#permalink]

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New post 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.

Answer: D.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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.


What are GMAT Club Tests?
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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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