dancinggeometry 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

Method 1:

Let A B C D E F G be the 7 people.

4 people have exactly 1 friend:

Grp1 of A and B . So each A and B will have one friend

Grp2 of C and D . So each C and D will have one friend

3 people have exactly 2 friends:

Grp3: E F G

Selecting 2 from 7 people: 7C2 = 21

Select two people who are not freinds ( this will be combinations of three)

Grp1+ Grp2= 2C1 * 2C1 = 4

Grp2+ Grp3= 2C1 * 3C1 = 6

Grp3+ Grp1= 2C1 * 3C1 = 6

Total= 6+6+4 = 16

Probability=16/21 (Answer - OA)

Method 2:

We are told that 4 people have exactly 1 friend. This would account for 2 friend relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 freinds. This would account for another 3 friend relationships (e.g. EF, EG, and FG). Thus, there are 5 total friend relationships in the group.

Additionally, there are (7 x 6)/2 = 21 different ways to chose two people (friend relationship) from the room.

Therefore, the probability that any 2 individuals in the group are friends is 5/21. The probability that any 2 individuals in the group are NOT friends = 1 – 5/21 = 16/21.

The correct answer is E

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