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

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Intern
Joined: 13 Jun 2005
Posts: 29
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Kudos [?]: 11 [0], given: 0

In a room filled with 7 people, 4 people have exactly 1 [#permalink]  29 Jun 2007, 14:09
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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

I don't have the answer, but am getting E :)
VP
Joined: 08 Jun 2005
Posts: 1147
Followers: 6

Kudos [?]: 128 [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

Intern
Joined: 28 Dec 2006
Posts: 13
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Kudos [?]: 0 [0], given: 0

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
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