It is currently 17 Oct 2017, 23:03

### 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
Intern
Joined: 13 Jun 2005
Posts: 28

Kudos [?]: 42 [0], given: 0

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

### Show Tags

29 Jun 2007, 15:09
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 (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?
5/21
3/7
4/7
5/7
16/21

I don't have the answer, but am getting E :)

Kudos [?]: 42 [0], given: 0

VP
Joined: 08 Jun 2005
Posts: 1144

Kudos [?]: 246 [0], given: 0

### Show Tags

30 Jun 2007, 00:22
(4/7)*(1/6) = 4/42 (choosing a person that has only 1 friend out of seven and choosing his friend out of 6)

(3/7)*(2/6) = 6/42 (choosing a person that has 2 friends out of seven and choosing his friend , he has two, out of 6)

4/42+6/42 = 10/42 = 5/21

but we need 1-(5/21) = 16/21

Kudos [?]: 246 [0], given: 0

Intern
Joined: 28 Dec 2006
Posts: 13

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

### Show Tags

30 Jun 2007, 00:49
i think the hardest part of this problem, is breaking down the relationships.

I added each person as a vertex A to G.

The only way to satisfy, the 3 people with only 2 friends is by drawing a triangle with three of the vertexes (A,B,C)

The only way to satisfy the 4 people with only 1 friend is by drawing 2 line segments (DE and FG).

That visual helps to determine the probability which is pretty straightforward. ..

chance of drawing a 2-friend person, 3/7 and then drawing a non-friend, 4/6
chance of drawing a 1-friend person, 4/7 and then drawing a non-friend, 5/6

(3/7)*(4/6) + (4/7)*(5/6) = 32/42 = 16/21

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

30 Jun 2007, 00:49
Display posts from previous: Sort by