In a room filled with 7 people, 4 people have exactly 1 : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 03:17

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In a room filled with 7 people, 4 people have exactly 1

Author Message
Manager
Joined: 25 May 2006
Posts: 227
Followers: 1

Kudos [?]: 60 [0], given: 0

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

### Show Tags

23 Jun 2006, 19:38
00:00

Difficulty:

(N/A)

Question Stats:

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

### HideShow timer Statistics

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

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. 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
_________________

Who is John Galt?

Manager
Joined: 01 Jun 2006
Posts: 140
Followers: 1

Kudos [?]: 5 [0], given: 0

### Show Tags

23 Jun 2006, 19:47
Pretend that we have A B C D E F G and AB,CD,EG,EF,GF are groups of friends.
So the needed prob must be 5/7C2 = 5/21
Manager
Joined: 25 May 2006
Posts: 227
Followers: 1

Kudos [?]: 60 [0], given: 0

### Show Tags

24 Jun 2006, 10:56
_________________

Who is John Galt?

Manager
Joined: 25 May 2006
Posts: 227
Followers: 1

Kudos [?]: 60 [0], given: 0

### Show Tags

24 Jun 2006, 17:50
Yeap, you got it Mr. Please show me the way, bcs I'm lost in this one
_________________

Who is John Galt?

Manager
Joined: 01 Jun 2006
Posts: 140
Followers: 1

Kudos [?]: 5 [0], given: 0

### Show Tags

24 Jun 2006, 18:48
[quote="X & Y"]mmmm noup. Please try again. [/quote]
It is my mistake. The explaination is right,huh?
But i calculated the prob that the chosen 2 people are friends.
It is obvious that the needed prob must be 1-5/21=16/21
E it is
Senior Manager
Joined: 07 Jul 2005
Posts: 404
Location: Sunnyvale, CA
Followers: 2

Kudos [?]: 13 [0], given: 0

### Show Tags

24 Jun 2006, 20:28
(E)

Total ways 2 individiuals are selected - 7C2 = 21

Total pair of friends -
4 friends have exactly 1 friend - 2 pairs
3 friends have exactly 2 - 3 pairs
= 5

prob. that 2 individuals are friends - 5/21
prob. not friends - 1-5/21 = 16/21 = (E)
Senior Manager
Joined: 16 Apr 2006
Posts: 276
Followers: 2

Kudos [?]: 181 [0], given: 2

### Show Tags

24 Jun 2006, 21:55
Quote:
(E)

Total ways 2 individiuals are selected - 7C2 = 21

Total pair of friends -
4 friends have exactly 1 friend - 2 pairs
3 friends have exactly 2 - 3 pairs
= 5

prob. that 2 individuals are friends - 5/21
prob. not friends - 1-5/21 = 16/21 = (E)

Thanks for explanation.
Senior Manager
Joined: 11 May 2006
Posts: 258
Followers: 2

Kudos [?]: 23 [0], given: 0

### Show Tags

25 Jun 2006, 10:21
mine approach is pretty much same as sgrover's.
25 Jun 2006, 10:21
Display posts from previous: Sort by