GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 13 Oct 2019, 23:34

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

M23-22

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58335
M23-22  [#permalink]

Show Tags

New post 16 Sep 2014, 01:19
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

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

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58335
Re M23-22  [#permalink]

Show Tags

New post 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);

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


Answer: C
_________________
Current Student
User avatar
Joined: 12 Nov 2015
Posts: 55
Location: Uruguay
Concentration: General Management
Schools: Goizueta '19 (A)
GMAT 1: 610 Q41 V32
GMAT 2: 620 Q45 V31
GMAT 3: 640 Q46 V32
GPA: 3.97
Re: M23-22  [#permalink]

Show Tags

New post 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?

:?
Intern
Intern
avatar
Joined: 22 Jul 2016
Posts: 1
Re: M23-22  [#permalink]

Show Tags

New post 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
Intern
avatar
Joined: 14 Oct 2015
Posts: 6
GMAT ToolKit User
Re: M23-22  [#permalink]

Show Tags

New post 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\(\frac{4}{5}\)
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58335
Re: M23-22  [#permalink]

Show Tags

New post 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.
_________________
Intern
Intern
avatar
Joined: 14 Oct 2015
Posts: 6
GMAT ToolKit User
Re: M23-22  [#permalink]

Show Tags

New post 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 \(\frac{4}{5}\)
Intern
Intern
avatar
Joined: 21 May 2017
Posts: 2
Re: M23-22  [#permalink]

Show Tags

New post 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.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58335
Re: M23-22  [#permalink]

Show Tags

New post 16 Jul 2017, 04:12
Senior Manager
Senior Manager
User avatar
S
Joined: 08 Jun 2015
Posts: 420
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33
Reviews Badge
Re: M23-22  [#permalink]

Show Tags

New post 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.
_________________
" The few , the fearless "
Intern
Intern
avatar
B
Joined: 27 Mar 2019
Posts: 38
Location: India
GMAT 1: 650 Q48 V32
GPA: 3
CAT Tests
Re: M23-22  [#permalink]

Show Tags

New post 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);

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


Answer: C


Bunuel can you please explain why the order does not matter?
Veritas Prep GMAT Instructor
User avatar
B
Joined: 01 May 2019
Posts: 50
Re: M23-22  [#permalink]

Show Tags

New post 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
Intern
avatar
B
Joined: 21 Mar 2015
Posts: 8
Re: M23-22  [#permalink]

Show Tags

New post 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
User avatar
V
Joined: 02 Sep 2009
Posts: 58335
Re: M23-22  [#permalink]

Show Tags

New post 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.
_________________
GMAT Club Legend
GMAT Club Legend
User avatar
D
Joined: 18 Aug 2017
Posts: 4976
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Re: M23-22  [#permalink]

Show Tags

New post 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
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
GMAT Club Bot
Re: M23-22   [#permalink] 04 Jul 2019, 07:56
Display posts from previous: Sort by

M23-22

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel






Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne