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In a room filled with 7 people

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by coolvishu11 » Sun May 10, 2009 5:14 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. 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

oa IS e
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by sureshbala » Sun May 10, 2009 8:17 pm
Let us name them as F1, F2, F3, F4, F5, F6 and F7.

Given that 4 of them are having exactly one friend.

So, F1 and F2 are friends and F3 and F4 are friends.

F1---F2

F3---F4


Also 3 of them exactly 2 friends.

i.e. F5, F6 and F7 are friends to each other.


i.e. F5 is a friend to F6 and F7

implies F6 is a friend to F5 and F7

and F7 is a friend to F5 and F6



Now if 2 members are selected,

P(that they are not friends) = 1 - P(that they are friends)

It is evident from the above data that the total number of ways of selecting 2 of them such that they are friends = 5.

i.e (F1,F2), (F3,F4), (F5,F6), (F5,F7) and (F6.F7)

Hence Prob that they are friends = 5/7C2 = 5/21.

So Prob that they are not friends = 1 - 5/21 = 16/21

(Folks, for clarity i have assumed named them and paired up as friends.)
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