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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]

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18 Nov 2007, 11:48
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?

a)5/21
b)3/7
c)4/7
d)5/7
e)16/21
CEO
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Concentration: Entrepreneurship, Other
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18 Nov 2007, 12:18
e

p=1-q=1-(4*1/6+3*2/6)/7=1-5/21=16/21
Manager
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18 Nov 2007, 17:54
walker wrote:
e

p=1-q=1-(4*1/6+3*2/6)/7=1-5/21=16/21

Walker, could you please explain in more detail?
VP
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18 Nov 2007, 18:18
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: 26 Sep 2007
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18 Nov 2007, 18:47
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

KillerSquirrel, thanks for the explanation.
Manager
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18 Nov 2007, 19:23
KillerSquirrel, thanks a lot - I was lost in this Q ...
18 Nov 2007, 19:23
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