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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In a room filled with 7 people, 4 people have exactly 1 sibl [#permalink]

Show Tags

14 Jul 2008, 17:24

1

This post received KUDOS

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

54% (02:49) correct
46% (01:34) wrong based on 52 sessions

HideShow timer Statistics

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (4-5-6).

Solution #1: # of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 people which are not siblings - \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings)+\(C^1_2*C^1_3\) (one from first pair of siblings*one from triple)+ \(C1^_2*C^1_3\)(one from second pair of siblings*one from triple) \(=4+6+6=16\).

\(P=\frac{16}{21}\)

Solution #2: # of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\);

Re: PS - probability that those two individuals are not friends? [#permalink]

Show Tags

14 Jul 2008, 18:06

1

This post received KUDOS

x-ALI-x wrote:

In a room filled 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 John is Peter's friend, Peter is John'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

Possible combinatios of selecting 2 people from 7 = 7C2 = 21

Possible Combinations of finding 2 people i.e a pair such that both are friends = 5 (I calculated this following way, 7 individuals named 1,2,..7; 4 have exactly one friend: 1-2, 3-4 3 have exactly 2 friends: 5-6, 5-7, 6-7 )

Probability of getting 2 people who are friends = 5/21

Probabilty of getting 2 people who are not friends = =1 - Probability of getting 2 people who are friends = 1- 5/21 = 16/21

Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

16 Sep 2013, 00:16

1

This post received KUDOS

First consider the relationships of individual A: AB, AC, AD, AE, AF, AG = 6 total. Then consider the relationships of individual B without counting the relationship AB that was already counted before: BC, BD, BE, BF, BG = 5 total. Continuing this pattern, we can see that C will add an additional 4 relationships, D will add an additional 3 relationships, E will add an additional 2 relationships, and F will add 1 additional relationship. Thus, there are a total of 6 + 5 + 4 + 3 + 2 + 1 = 21 total relationships between the 7 individuals.

We are told that 4 people have exactly 1 friend. This would account for 2 "friendship" relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 friends. This would account for another 3 "friendship" relationships (e.g. EF, EG, and FG). Thus, there are 5 total "friendship" relationships in the group. The probability that any 2 individuals in the group are friends is 5/21. The probability that any 2 individuals in the group are NOT friends = 1 – 5/21 = 16/21. The correct answer is E.

Re: PS - probability that those two individuals are not friends? [#permalink]

Show Tags

14 Jul 2008, 17:52

In a room filled 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 John is Peter's friend, Peter is John'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

lets see total possibility = 7C2 =21

so out of the 4,.we have 4c1 ways to pick 1 from this group of 4 and 3c1 from out of the other group 3c1..

Re: PS - probability that those two individuals are not friends? [#permalink]

Show Tags

14 Jul 2008, 19:41

the question is quite simple the 4 guys who ONLY know 1 person other than themselves is one group of 4 people..then out of the 7 only the other 3 know each other...but these 2 groups dont know anyone from outside of their group..

so the question is asking pick one guy from 1 group (i.e the 4 guys who only have 1 friend) and one guy from group 2 (i.e the group of 3 socialites)..

so basically you have 4c1 and 3c1 possibilites picking 2 people..

total possibility of picking any 2 people is 7c2=27..

alpha_plus_gamma wrote:

x-ALI-x wrote:

In a room filled 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 John is Peter's friend, Peter is John'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

Possible combinatios of selecting 2 people from 7 = 7C2 = 21

Possible Combinations of finding 2 people i.e a pair such that both are friends = 5 (I calculated this following way, 7 individuals named 1,2,..7; 4 have exactly one friend: 1-2, 3-4 3 have exactly 2 friends: 5-6, 5-7, 6-7 )

Probability of getting 2 people who are friends = 5/21

Probabilty of getting 2 people who are not friends = =1 - Probability of getting 2 people who are friends = 1- 5/21 = 16/21

Re: PS - probability that those two individuals are not friends? [#permalink]

Show Tags

14 Jul 2008, 20:10

fresinha12 wrote:

the question is quite simple the 4 guys who ONLY know 1 person other than themselves is one group of 4 people..then out of the 7 only the other 3 know each other...but these 2 groups dont know anyone from outside of their group..

so the question is asking pick one guy from 1 group (i.e the 4 guys who only have 1 friend) and one guy from group 2 (i.e the group of 3 socialites)..

so basically you have 4c1 and 3c1 possibilites picking 2 people..

total possibility of picking any 2 people is 7c2=27..

you can pick two guys from group 1 and still they wont be fried with eachother 1234 is the group of 4 guys, 1 is fried with 2 and 3 is frid with 4.... this doesnt mean that 1 is friend with 3

so you can select (1-3, 1-4, 2-3, 2-4) from the same group...... so you have to add 4 to 12 for total fav

the answer should be 16/21

gmatclubot

Re: PS - probability that those two individuals are not friends?
[#permalink]
14 Jul 2008, 20:10

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...