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

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Director
Joined: 27 Jun 2005
Posts: 513
Location: MS
Followers: 2

Kudos [?]: 36 [0], given: 0

In a room filled with 7 people, 4 people have exactly 1 [#permalink]  13 Feb 2006, 10:22
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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?

5/21
3/7
4/7
5/7
16/21
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
Followers: 3

Kudos [?]: 40 [0], given: 0

Is it 16/21?

There can be only 2 groups possible among 7 of them:
Group1 : ABCD - Have of them exactly 1 friends
Group2 : EFG - Have exactly 2 friends

P(Both NOT friends) = 1 - P(Both friends) = 1 - [ P(two friends are from Group1) + P (two of them are from Group2)]

P(two of them have exactly 1 friend) = 4/7*1/6 -- Choose any of the 4 friends from Group1 and then choose HIS only freind

P (two of them have exactly 2 friends) = 3/7*2/6 -- Choose any of the 3 friends from Group2 and then choose one of his 2 freinds

P(Both NOT freinds) = 1 - { 4/42+6/42} = 16/21.

I may be wrong....
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Director
Joined: 27 Jun 2005
Posts: 513
Location: MS
Followers: 2

Kudos [?]: 36 [0], given: 0

giddi77 wrote:
Is it 16/21?

There can be only 2 groups possible among 7 of them:
Group1 : ABCD - Have of them exactly 1 friends
Group2 : EFG - Have exactly 2 friends

P(Both NOT friends) = 1 - P(Both friends) = 1 - [ P(two friends are from Group1) + P (two of them are from Group2)]

P(two of them have exactly 1 friend) = 4/7*1/6 -- Choose any of the 4 friends from Group1 and then choose HIS only freind

P (two of them have exactly 2 friends) = 3/7*2/6 -- Choose any of the 3 friends from Group2 and then choose one of his 2 freinds

P(Both NOT freinds) = 1 - { 4/42+6/42} = 16/21.

I may be wrong....

you got it right OA is 16/21
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