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.