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

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

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17 Jul 2006, 16:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Can someone please show me a quick way to approach this question. I figured it out but it's taken me way too long..

SQ
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17 Jul 2006, 18:26
Total selections = 21
Consider A B C D E F G
There can be 5 set of friends
consider AB CD EF FG GE
prob of choosing friends = 5/21
not choosing friends = 16/21 (E)
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17 Jul 2006, 18:32
Let's call the seven people ABCD EFG (ABCD have only one friend, EFG have two friends).

Assumption: If X is a friend of Y, then Y is a friend of X ("a friendship") for any X,Y (element of ABCDEFG, x is not y).

If this assumption is correct, then it is pretty easy to calculate the probabilities:

Consider the first person, that is selected randomly.

In 4 of 7 cases, this person has only friend.
Hence the probability that the other person isn't his friend is 4/7*5/6=20/42.

In 3 of 7 cases, this person has two friends.
Hence the probability that the other person isn't his friend is 3/7*4/6=12/42.

Putting both cases together yields: 20/42+12/42=32/42=16/21.

Therefore E.
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18 Jul 2006, 03:26
Nicely done!

Or, think of three groups of people

Group 1: abc
Group 2: de
Group 3: fg

Number of ways of choosing 2 people 6C2=21
Number of ways of choosing 2 friends 3+1+1=5

Therefore, the probability that the two chosen people are not friends is (21-5)/21= 16/21
18 Jul 2006, 03:26
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# In a room filled with 7 people, 4 people have exactly 1

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