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

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Manager
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In a room filled with 7 people, 4 people have exactly 1 [#permalink] New post 16 Sep 2008, 07:42
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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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Re: Zumit PS 022 [#permalink] New post 16 Sep 2008, 12:44
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let a b c d e f and g represent the people andlet the following pairs depict friends:

a-b, c-d, e-f-g (e-f-g should be viewed in a triangular arrangement such that e-f, e-g, g-f are the friends)
Let there be 3 groups: Gp1: a-b; Gp2: c-d; Gp3: e-f-g

Gp1 + Gp2: (2C1*2C1)/7C2 = 8/42
Gp1 + Gp3: (2C1*3C1)/7C2 = 12/42
Gp2 + Gp3: (2C1*3C1)/7C2 = 12/42
Therefore, total prob. = 32/42 = 16/21
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Re: Zumit PS 022 [#permalink] New post 16 Sep 2008, 15:29
I get 4/7

lets see..4 people know exactly 1 friend..which means.. there exactly 4 people who have 1 other friend..leaving behind only 3..

so total probability of picking 2 people is 7c2

prob that you pick 4c1 and 3c1 as the two people..you get 4/7
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Re: Zumit PS 022 [#permalink] New post 22 Sep 2011, 19:09
dancinggeometry wrote:
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


Method 1:
Let A B C D E F G be the 7 people.
4 people have exactly 1 friend:
Grp1 of A and B . So each A and B will have one friend
Grp2 of C and D . So each C and D will have one friend

3 people have exactly 2 friends:
Grp3: E F G

Selecting 2 from 7 people: 7C2 = 21
Select two people who are not freinds ( this will be combinations of three)
Grp1+ Grp2= 2C1 * 2C1 = 4
Grp2+ Grp3= 2C1 * 3C1 = 6
Grp3+ Grp1= 2C1 * 3C1 = 6

Total= 6+6+4 = 16
Probability=16/21 (Answer - OA)

Method 2:
We are told that 4 people have exactly 1 friend. This would account for 2 friend relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 freinds. This would account for another 3 friend relationships (e.g. EF, EG, and FG). Thus, there are 5 total friend relationships in the group.

Additionally, there are (7 x 6)/2 = 21 different ways to chose two people (friend relationship) from the room.

Therefore, the probability that any 2 individuals in the group are friends is 5/21. The probability that any 2 individuals in the group are NOT friends = 1 – 5/21 = 16/21.

The correct answer is E
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Re: Zumit PS 022   [#permalink] 22 Sep 2011, 19:09
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