BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

MGMAT - probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pm
Location: Silicon valley, California
Thanked: 30 times
Followed by:1 members

MGMAT - probability

by jayhawk2001 » Sun May 13, 2007 3:06 pm
Interesting probability question. OA after a few reply.

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
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 519
Joined: Sat Jan 27, 2007 7:56 am
Location: India
Thanked: 31 times

Neat Problem

by Neo2000 » Sun May 13, 2007 11:35 pm
Pretty interesting problem I only hope I have the answer correct

16/21
Please see attached image [/img]


4People who know exactly 1

1->5
2->6
3->7
4->3

3People who know exactly 2
3->4,7
2->6,1
1->5,2

Now, to pick 2 people such that they are not friends:

Picking 2 people from 7 in 7C2 = 21

Picking 2people such that they do not know each other

Ways of picking people such that they know each other
1 (2,5)
2 (6) since 2 knowing 1 is same as 1 knowing 2
3 (4,7)
Total = 5
Therefore Number of ways of picking people such that they do not know each other = 21 - 5 = 16

Total Probability = 16/21
Attachments
Soln.JPG
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Mon May 07, 2007 12:36 pm

by chitoshia » Mon May 14, 2007 4:59 am
Choosing between any of the groups in the attached image, and the rest of the numbers are the non-friends one must choose from

The probability:
=((2C1 X 5C1) + (2C1 x C1) + (3C1 x 4C1))/(7C1 x 6C1)
=32/42
=16/21

Hence (E)

Wish I could explain more clearly, nice question though.

Regards
Attachments
friends.JPG
The relationship
Top
User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pm
Location: Silicon valley, California
Thanked: 30 times
Followed by:1 members

by jayhawk2001 » Mon May 14, 2007 7:00 am
Both approaches are correct. OA is indeed E
Top
Post new topic Post Reply
• Page 1 of 1