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:

MGMAT Test1:#29 In a room filled with 7 people, 4 people [#permalink]
29 Mar 2008, 11:55

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

MGMAT Test1:#29

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

I know OA and OE but didn't get it ( it is confusing for me). I appreciate different approaches and views Explanations plz, thx.

Re: PS_Probability_friendship [#permalink]
29 Mar 2008, 15:45

I will solve it using Combination approach. We have 4 people with 1 friend, so total ways for that = 4C1 = 4 We have 3 people with 2 friend, so total ways for that = 3C2 = 3

Since friendship is mutually exclusive so people above two group will not be friends of each others.

Total ways of selecting 2 people from 7 people = 7C2 = 21

Re: PS_Probability_friendship [#permalink]
29 Mar 2008, 17:33

I agree with abhijit_sen's answer. Here's another way to look at things:

It may be easier to understand this by drawing arranging 7 letters A to G in a circle. You'll see that there is no way to achieve the "4 have exactly 1 friend and 3 have exactly 2 friends" requirement without drawing a triangle among 3 people and using 2 lines to pair up the remaining 4 people. For the sake of this argument, let ABC be the triangle, DE be a pair, and FG the last pair.

Now find the number of ways of there not being a pair of friends selected. Let's say A is chosen first, there are only 4 people that can be chosen that won't a friend. This also applies for B and C, so altogether, there are 12 ways for this to happen.

(Notice that we don't do need to do the reverse where we choose D,E,F or G first because the would only end up choosing A, B, or C. We've already handled the case and the order doesn't matter)

Total ways = 7C2

Hence, 12/21 is the probability which simplifies to 4/7. Answer C

Re: PS_Probability_friendship [#permalink]
30 Mar 2008, 00:47

agree with Walker solution, it just says friend in a room, not necessary to break out into 2 grps, i think that is the confusing part. C is only correct when we make above assumption

Re: PS_Probability_friendship [#permalink]
30 Mar 2008, 09:01

Hi, this is a long routed solution but it may help to see the logic. Lets label the members as A,B,C,D,E,F,G We begin by counting the relationships

A AB,AC,AD,AE,AF,AG =6 total B BC,BD,BE,BF,BG =5 total (we do not count BA because the relationship is mutual and we would be counting it twice) C CD,CE,CF,CG = 4 total D DE,DF,DG =3 E EF,EG =2 F FG =1 Therefore the total number of relationships is 6+5+4+3+2+1 =21 Criteria 1 If 4 people have exactly one friend then the number of relationships is 2 eg AB ,CD (A is a friend of B and B is a friend of A (1 relationship between 2 people), C is a friend of D and D is a friend of C (2nd relationship between 2 other people)Therefore we have 2 relationships between 4 people.=2

Criteria 2

3 people have exactly 2 friends in the room

CD,CE,DE (C is a friend of D and E (1 relationship), D is a friend of E and C, (2nd relationship) E is a friend of C and D(3rd relationship)) So we have 3 relationships in total that satisfy the above criteria =3

The probability that the two random individuals are friends =(3+2)/21 =5/21 The probability that the two random individuals chosen are not friends = 1- (5/21)