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

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

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New post 22 May 2007, 00:24
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

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New post 22 May 2007, 01:29
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 :)

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New post 22 May 2007, 01:43
(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

the answer is (E)

:-D

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New post 23 May 2007, 08:03
AdrianG 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

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New post 23 May 2007, 08:06
ronron wrote:
AdrianG 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

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New post 23 May 2007, 08:18
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

the answer is (E)

:-D


thanks, KillerSquirrel

fast way to solve it.

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New post 24 May 2007, 01:13
good job Killer Squirrel

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New post 24 May 2007, 02:22
AdrianG wrote:
ronron wrote:
AdrianG 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

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New post 24 May 2007, 02:41
ronron wrote:
AdrianG wrote:
ronron wrote:
AdrianG 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

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 [#permalink]

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New post 24 May 2007, 06:11
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

the answer is (E)

:-D


Thanks a lot! Wonderful and the best explanation than anyone else!

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New post 24 May 2007, 10:14
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

ANSWER= E

WHAT IS THE OA

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 [#permalink]

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New post 24 May 2007, 18:42
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

ANSWER= E

WHAT IS THE OA


Your answer is correct!

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New post 18 Aug 2007, 17:07
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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  [#permalink] 18 Aug 2007, 17:07
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