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Friends in a room

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Friends in a room

by csandeepreddy » Mon Oct 20, 2008 7:19 am
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: Friends in a room

by sudhir3127 » Mon Oct 20, 2008 7:23 am
csandeepreddy 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
Sandeep,
Please use the search option before starting a new thread.. this Question has been answered in the past..

The total no. of selection of 2 persons from 7 = 7c2 =21

let the number of persons in the room be 1, 2, 3, 4, 5, 6 and 7

possible friendship combinations is 1-5 , 2-5 , 3-6 , 4-7, 6-7

There will be 5 pair of friends which will satisfy the given condition so probability of selcting a pair of friends out of 21 pairs = 5/21

prob. of not selecting friends= 1- 5/21 =16/21
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by aditi_bc » Tue Oct 21, 2008 9:34 am
i don't understand how to picked these combinations?
"possible friendship combinations is 1-5 , 2-5 , 3-6 , 4-7, 6-7 "

Could u pl explain further?
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by cramya » Tue Oct 21, 2008 2:42 pm
Answering for Sudhir hoping he woudnt mind

It doesnt have to be exactly 1-5 , 2-5 , 3-6 , 4-7, 6-7

It could be 1-3 , 2-3 , 5-6 , 4-7, 6-7
i.e 1,2,4,5 have exactly 1 friend whereas 3,6,7 have exactly 2 friends
(6-7 works since the friendship is mutual so 6's 2 friend condn and 7's 2 friend condition is satisfied by this single pairing given 5-6 and 4-7 exists)

Hope this helps!
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