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

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New post 23 Jun 2006, 19:38
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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. 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
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New post 23 Jun 2006, 19:47
Pretend that we have A B C D E F G and AB,CD,EG,EF,GF are groups of friends.
So the needed prob must be 5/7C2 = 5/21
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New post 24 Jun 2006, 10:56
mmmm noup. Please try again. :)
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New post 24 Jun 2006, 17:50
Yeap, you got it Mr. Please show me the way, bcs I'm lost in this one :?
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New post 24 Jun 2006, 18:48
[quote="X & Y"]mmmm noup. Please try again. :)[/quote]
It is my mistake. The explaination is right,huh?
But i calculated the prob that the chosen 2 people are friends.
It is obvious that the needed prob must be 1-5/21=16/21
:-D E it is
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New post 24 Jun 2006, 20:28
(E)

Total ways 2 individiuals are selected - 7C2 = 21

Total pair of friends -
4 friends have exactly 1 friend - 2 pairs
3 friends have exactly 2 - 3 pairs
= 5

prob. that 2 individuals are friends - 5/21
prob. not friends - 1-5/21 = 16/21 = (E)
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New post 24 Jun 2006, 21:55
Quote:
(E)

Total ways 2 individiuals are selected - 7C2 = 21

Total pair of friends -
4 friends have exactly 1 friend - 2 pairs
3 friends have exactly 2 - 3 pairs
= 5

prob. that 2 individuals are friends - 5/21
prob. not friends - 1-5/21 = 16/21 = (E)


Thanks for explanation.
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New post 25 Jun 2006, 10:21
mine approach is pretty much same as sgrover's.
  [#permalink] 25 Jun 2006, 10:21
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