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

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Director
Joined: 03 Sep 2006
Posts: 884
Followers: 6

Kudos [?]: 288 [0], given: 33

In a room filled with 7 people, 4 people have exactly 1 [#permalink]  21 May 2007, 23:24
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
Manager
Joined: 09 Nov 2006
Posts: 128
Followers: 1

Kudos [?]: 3 [0], given: 0

I imagine the "friendship scheme" as this:

1 and 2 are friends (each has exactly 1 friend)
3 and 4 are friends (each has exactly 1 friend)
5, 6 and 7 are friends (each has exactly 2 friends)

Total possible number of outcomes of picking 2 random ppl from 7 is 7!/5!2! = 21

If 1 is randomly picked, than there is only 1 outcome that another randomly picked will be his friend.
The same goes for number 2,3 and 4
For numbers 5,6 and 7 there are 2 outcomes respectivly. In total there are 10 "favourable" outcomes. Means that 11 are "unfavourable". But there is no 11/21 among answers:) So, I must have done a mistake in my last part of calculations
VP
Joined: 08 Jun 2005
Posts: 1146
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Kudos [?]: 136 [0], given: 0

(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

Manager
Joined: 22 May 2007
Posts: 115
Followers: 1

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

5, 6 and 7 are friends (each has exactly 2 friends)

This is not always true

5-6-7

6 can be friends with both 5 and 7 but 5 n doesn't have to be friends with 7
Manager
Joined: 09 Nov 2006
Posts: 128
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Kudos [?]: 3 [0], given: 0

ronron wrote:
5, 6 and 7 are friends (each has exactly 2 friends)

This is not always true

5-6-7

6 can be friends with both 5 and 7 but 5 n doesn't have to be friends with 7

Actually 5 has to be friend of 7 also. He can't be friend of 1,2,3 and 4
Senior Manager
Joined: 03 May 2007
Posts: 272
Followers: 1

Kudos [?]: 8 [0], given: 0

KillerSquirrel wrote:
(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

thanks, KillerSquirrel

fast way to solve it.
Manager
Joined: 13 Feb 2007
Posts: 63
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Kudos [?]: 1 [0], given: 0

Prob [#permalink]  24 May 2007, 00:13
good job Killer Squirrel
Manager
Joined: 22 May 2007
Posts: 115
Followers: 1

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

ronron wrote:
5, 6 and 7 are friends (each has exactly 2 friends)

This is not always true

5-6-7

6 can be friends with both 5 and 7 but 5 n doesn't have to be friends with 7

Actually 5 has to be friend of 7 also. He can't be friend of 1,2,3 and 4

another possible friends' scheme:

1-2-3-4 (1 with 2, 2 with 1 & 3, 3 with 2 & 4, 3 with 4)

5-6-7 (5 with 6, 6 with 5 & 7, 7 with 6)

so they don't always have to be friends
Manager
Joined: 09 Nov 2006
Posts: 128
Followers: 1

Kudos [?]: 3 [0], given: 0

ronron wrote:
ronron wrote:
5, 6 and 7 are friends (each has exactly 2 friends)

This is not always true

5-6-7

6 can be friends with both 5 and 7 but 5 n doesn't have to be friends with 7

Actually 5 has to be friend of 7 also. He can't be friend of 1,2,3 and 4

another possible friends' scheme:

1-2-3-4 (1 with 2, 2 with 1 & 3, 3 with 2 & 4, 3 with 4)

5-6-7 (5 with 6, 6 with 5 & 7, 7 with 6)

so they don't always have to be friends

Ronon, in your scheme 4 ppl (exactly 1,2,3 and 6) have 2 friends while the stem says that only 3 ppl have exactly 2 friends
Director
Joined: 03 Sep 2006
Posts: 884
Followers: 6

Kudos [?]: 288 [0], given: 33

KillerSquirrel wrote:
(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

:-D

Thanks a lot! Wonderful and the best explanation than anyone else!
Manager
Joined: 14 Mar 2007
Posts: 236
Followers: 1

Kudos [?]: 1 [0], given: 0

you can make the 7 friends:

A B C D E F G

4= A B C D
1 FRIEND MUTUAL = AB CD

3= E F G
2 FRIENDS MUTUAL = EF EG FG

SO FROM THE ABOVE= 2/4*2/3 = 1/3

TOTAL FRIENDS = 5
TOTAL PEOPLE = 7

SO, 5/7

PROB OF YES = 1/3*5/7 = 5/21

PROB OF NOT = 16/21

WHAT IS THE OA
Director
Joined: 03 Sep 2006
Posts: 884
Followers: 6

Kudos [?]: 288 [0], given: 33

andrehaui wrote:
you can make the 7 friends:

A B C D E F G

4= A B C D
1 FRIEND MUTUAL = AB CD

3= E F G
2 FRIENDS MUTUAL = EF EG FG

SO FROM THE ABOVE= 2/4*2/3 = 1/3

TOTAL FRIENDS = 5
TOTAL PEOPLE = 7

SO, 5/7

PROB OF YES = 1/3*5/7 = 5/21

PROB OF NOT = 16/21

WHAT IS THE OA

Director
Joined: 06 Sep 2006
Posts: 745
Followers: 1

Kudos [?]: 18 [0], given: 0

Out of 7 people (A, B, C, D, E, F, G)

ABCD has only one friends. May be A--B and C--D...

EFG has two friends; E is friends with F and G;....

Probability For People with only 1 friend:
4/7 * 1/6 [once we picked A; in the second go we can only pick B]
=2/21

Prob. for people with 2 friends:
3/7 * 2/6 [If we pick E, for next go we can pick either F or G hence 2]

= 3/21

Probability of picking two people who ARE friends = 2/21 + 3/21 = 5/21

two people who are NOT friends = 1 - 5/21 = 16/21.
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