Find all School-related info fast with the new School-Specific MBA Forum

It is currently 01 Sep 2014, 20:29

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

In a room with 7 people, 4 people have exactly 1 friend in

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
3 KUDOS received
Intern
Intern
avatar
Joined: 13 Jun 2007
Posts: 48
Followers: 1

Kudos [?]: 3 [3] , given: 0

In a room with 7 people, 4 people have exactly 1 friend in [#permalink] New post 08 Dec 2007, 12:00
3
This post received
KUDOS
4
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

56% (03:38) correct 44% (02:23) wrong based on 116 sessions
In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
[Reveal] Spoiler: OA
Kaplan Promo CodeKnewton GMAT Discount CodesVeritas Prep GMAT Discount Codes
3 KUDOS received
Director
Director
User avatar
Joined: 12 Jul 2007
Posts: 867
Followers: 12

Kudos [?]: 196 [3] , given: 0

GMAT Tests User
 [#permalink] New post 08 Dec 2007, 12:09
3
This post received
KUDOS
2
This post was
BOOKMARKED
(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.

20/42 + 12/42
32/42 = 16/21

E. 16/21
VP
VP
avatar
Joined: 22 Nov 2007
Posts: 1102
Followers: 6

Kudos [?]: 122 [0], given: 0

GMAT Tests User
Re: [#permalink] New post 16 Jan 2008, 22:04
eschn3am wrote:
(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.

20/42 + 12/42
32/42 = 16/21

E. 16/21


excuse me, can you explain me this passage in detail? why do we have to sum probabilities up? don't we need to multiply them? we need that both conditions apply. thanks a lot
Director
Director
User avatar
Joined: 12 Jul 2007
Posts: 867
Followers: 12

Kudos [?]: 196 [0], given: 0

GMAT Tests User
Re: PS: Combinatoric [#permalink] New post 17 Jan 2008, 04:51
We don't need to multiply them together because we're only choosing one pair of people.

Image = probability of choosing 2 people who aren't friends out of the 4 with one friend
Image = probability of choosing 2 people who aren't friends out of the 3 with two friends

Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.

7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way
VP
VP
avatar
Joined: 22 Nov 2007
Posts: 1102
Followers: 6

Kudos [?]: 122 [0], given: 0

GMAT Tests User
Re: PS: Combinatoric [#permalink] New post 18 Jan 2008, 04:17
eschn3am wrote:
We don't need to multiply them together because we're only choosing one pair of people.

Image = probability of choosing 2 people who aren't friends out of the 4 with one friend
Image = probability of choosing 2 people who aren't friends out of the 3 with two friends

Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.

7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way


right....let's solve it with combs:

How many different subsets of two people can be formed from the seven people? 7C2 = 21


How many of these subsets contain two friends? please give me a hand in doing this...
4 KUDOS received
Director
Director
avatar
Joined: 01 Jan 2008
Posts: 629
Followers: 3

Kudos [?]: 130 [4] , given: 1

GMAT Tests User
Re: PS: Combinatoric [#permalink] New post 18 Jan 2008, 07:00
4
This post received
KUDOS
marcodonzelli wrote:
right....let's solve it with combs:

How many different subsets of two people can be formed from the seven people? 7C2 = 21

How many of these subsets contain two friends? please give me a hand in doing this...


you are right that there are 21 possible ways of choosing a pair. I suggest visualizing a graph of friendships. 1 and 2 are friends; 3 and 4 are friends; 5, 6 and 7 are friends (3 pairs here: 5-6, 6-7, 5-7). Therefore, there are 5 total pairs of friends. Then probability of choosing a pair who are not friends is 1-5/21 = 16/21. I hope that helps.
Expert Post
1 KUDOS received
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3571
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 358

Kudos [?]: 1759 [1] , given: 358

GMAT ToolKit User GMAT Tests User Premium Member
Re: PS: Combinatoric [#permalink] New post 19 Jan 2008, 03:27
1
This post received
KUDOS
Expert's post
an unusual way: :)

p=1-\frac{\frac{4*C_1^6+3*C_2^6}{2}}{C_2^7}=1-\frac{5}{21}=\frac{16}{21}
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Manager
avatar
Joined: 09 Apr 2008
Posts: 53
Followers: 0

Kudos [?]: 4 [0], given: 4

Re: PS: Combinatoric [#permalink] New post 13 Aug 2008, 23:30
walker wrote:
an unusual way: :)

p=1-\frac{\frac{4*C_1^6+3*C_2^6}{2}}{C_2^7}=1-\frac{5}{21}=\frac{16}{21}


Can you explain this a little more please? I understand the denominator - pick two people from 7. I think the numerator is the different combinations of people who ARE friends. But how did you get this?
Expert Post
1 KUDOS received
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3571
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 358

Kudos [?]: 1759 [1] , given: 358

GMAT ToolKit User GMAT Tests User Premium Member
Re: PS: Combinatoric [#permalink] New post 14 Aug 2008, 11:13
1
This post received
KUDOS
Expert's post
4*C_1^6 - for each person out of 4 we have only one friend out of remained people (7-1=6)
3*C_2^6 - for each person out of 3 we have two friends out of remained people (7-1=6)

\frac{4*C_1^6+3*C_2^6}{2} - excluding double counting.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Senior Manager
Senior Manager
User avatar
Joined: 07 Jan 2008
Posts: 418
Followers: 3

Kudos [?]: 64 [0], given: 0

GMAT Tests User
Re: PS: Combinatoric [#permalink] New post 19 Aug 2008, 22:54
my way:

p= 1-\frac{1+1+3}{C_2^7}= \frac{16}{21}
1 KUDOS received
Manager
Manager
avatar
Joined: 27 Oct 2008
Posts: 186
Followers: 1

Kudos [?]: 77 [1] , given: 3

GMAT Tests User
Re: PS: Combinatoric [#permalink] New post 27 Sep 2009, 11:26
1
This post received
KUDOS
In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
1)5/21
2)3/7
3)4/7
4)5/7
5)16/21

Soln Assuming the people to be A,B,C,D,E,F,G

We can arrange in this way such that 3 people have two friends while 4 have 1 friend
A B
B A
C D E
D C E
E C D
F G
G F

First column represents the 7 people, second and third columns represent their friends.
Thus A,B,F,G each have one friend while C,D,E have two friends

Probability when two people are chosen where they are not friends is
= (5 + 5 + 2 + 2 + 2)/7C2
= 16/21

Ans is option 5
Manager
Manager
avatar
Joined: 09 Aug 2009
Posts: 56
Followers: 1

Kudos [?]: 4 [0], given: 1

Re: PS: Combinatoric [#permalink] New post 06 Oct 2009, 23:01
Only one friend
A<->B
C<->D
two friends
E<->F
E<->G
F<->G

so total possibility 21
=4*3+4/21
=16/21

good question........
Senior Manager
Senior Manager
User avatar
Joined: 17 Dec 2012
Posts: 394
Location: India
Followers: 12

Kudos [?]: 175 [0], given: 9

Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink] New post 23 Apr 2013, 05:47
alexperi wrote:
In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
1)5/21
2)3/7
3)4/7
4)5/7
5)16/21


1. We will find the reverse case, i.e., 2 people who are selected, are friends.
2. There are two groups here. People with 1 friend and people with 2 friends.
3. Friends can be selected as follows: (i) Among the 4 people each with just 1 friend, there will be 2 pairs of friends. (ii) Among the 3 people each with 2 friends, there will be 3 pairs of friends.
4. So 5 cases of friends can be selected.
5. The total number of ways of selecting 2 people is7C2 = 21
6. The number of cases when the 2 selected people are not friends = 21-5 = 16
7. Probability that the two slected people are not friends= 16/21
The answer is therefore E.
_________________

Srinivasan Vaidyaraman
Sravna Test Prep
http://www.sravna.com/courses.php

Classroom Courses in Chennai
Online and Correspondence Courses

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25237
Followers: 3429

Kudos [?]: 25228 [0], given: 2702

Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink] New post 23 Apr 2013, 05:51
Expert's post
alexperi wrote:
In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21


OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: In a room with 7 people, 4 people have exactly 1 friend in   [#permalink] 23 Apr 2013, 05:51
    Similar topics Author Replies Last post
Similar
Topics:
In a room filled with 7 people, 4 people have exactly 1 Hermione 6 04 Nov 2006, 02:28
In a room filled with 7 people, 4 people have exactly 1 Paayal 12 27 Sep 2006, 08:31
In a room filled with 7 people, 4 people have exactly 1 shehreenquayyum 3 17 Jul 2006, 16:47
In a room filled with 7 people, 4 people have exactly 1 X & Y 7 23 Jun 2006, 19:38
In a room filled with 7 people, 4 people have exactly 1 cool_jonny009 2 13 Feb 2006, 10:22
Display posts from previous: Sort by

In a room with 7 people, 4 people have exactly 1 friend in

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.