Pretty interesting problem I only hope I have the answer correct

16/21
Please see attached image [/img]


4People who know exactly 1

1->5
2->6
3->7
4->3

3People who know exactly 2
3->4,7
2->6,1
1->5,2

Now, to pick 2 people such that they are not friends:

Picking 2 people from 7 in 7C2 = 21

Picking 2people such that they do not know each other

Ways of picking people such that they know each other
1 (2,5)
2 (6) since 2 knowing 1 is same as 1 knowing 2
3 (4,7)
Total = 5
Therefore Number of ways of picking people such that they do not know each other = 21 - 5 = 16

Total Probability = 16/21

Choosing between any of the groups in the attached image, and the rest of the numbers are the non-friends one must choose from

The probability:
=((2C1 X 5C1) + (2C1 x C1) + (3C1 x 4C1))/(7C1 x 6C1)
=32/42
=16/21

Hence (E)

Wish I could explain more clearly, nice question though.

Regards

Both approaches are correct. OA is indeed E