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VERY HARD COMBINATIONS - FRIENDS

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VERY HARD COMBINATIONS - FRIENDS

by canuckclint » Sat Oct 25, 2008 5:52 pm
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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by earth@work » Sat Oct 25, 2008 6:32 pm
Let A1 A2 A3 A4 have exactly 1 friend and B1 B2 B3 have exactly 2 friends
we will first find prob(friends together)
5 outcomes of freinds together r possible
A1B1, A2B1, A3B2, A4B3, B3B4…it can be any other combinations as well..like A1B3,A2B2, A3B3, A4B1,B1B2… =5 fav outcomes(that friends r picked).
Total outcomes = 7C2
P(2 friends picked) = 5/7C2…p(friends not picked) = 1-5/7c2=16/21

this has been discussed few days back, check this link for other explanation
https://www.beatthegmat.com/friends-in-a ... 21125.html
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