Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 27 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Jul 28 07:00 PM EDT  08:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Sunday, July 28th at 7 PM EDT
Author 
Message 
TAGS:

Hide Tags

Director
Joined: 07 Jun 2004
Posts: 577
Location: PA

If two different numbers are randomly selected from set { 1,
[#permalink]
Show Tags
Updated on: 13 Dec 2012, 02:07
Question Stats:
75% (00:47) correct 25% (03:54) wrong based on 25 sessions
HideShow timer Statistics
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 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
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.




Math Expert
Joined: 02 Sep 2009
Posts: 56370

If two different numbers are randomly selected from set { 1,
[#permalink]
Show Tags
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: \(P(of \ event \ X)=1P(of \ opposite \ event \ X)\). 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.
_________________




Manager
Joined: 06 Jul 2010
Posts: 68

Re: Probability clarification
[#permalink]
Show Tags
23 Jul 2010, 09:15
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: 577
Location: PA

Re: Probability clarification
[#permalink]
Show Tags
23 Jul 2010, 09:46
you are right thanks!
the key is "different" numbers in the Q
i missed that



Intern
Joined: 18 Mar 2012
Posts: 13

Re: Probability clarification
[#permalink]
Show Tags
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: 34
Concentration: Strategy, Marketing
GMAT Date: 06182013
GPA: 3.98

Re: Probability clarification
[#permalink]
Show Tags
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



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6991
Location: United States (CA)

Re: If two different numbers are randomly selected from set { 1,
[#permalink]
Show Tags
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.
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Intern
Joined: 27 Mar 2018
Posts: 11

Re: If two different numbers are randomly selected from set { 1,
[#permalink]
Show Tags
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: \(P(of \ event \ X)=1P(of \ opposite \ event \ X)\). 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: 56370

Re: If two different numbers are randomly selected from set { 1,
[#permalink]
Show Tags
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: \(P(of \ event \ X)=1P(of \ opposite \ event \ X)\). 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.
_________________




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






