i think the hardest part of this problem, is breaking down the relationships.
I added each person as a vertex A to G.
The only way to satisfy, the 3 people with only 2 friends is by drawing a triangle with three of the vertexes (A,B,C)
The only way to satisfy the 4 people with only 1 friend is by drawing 2 line segments (DE and FG).
That visual helps to determine the probability which is pretty straightforward. ..
chance of drawing a 2-friend person, 3/7 and then drawing a non-friend, 4/6
chance of drawing a 1-friend person, 4/7 and then drawing a non-friend, 5/6
(3/7)*(4/6) + (4/7)*(5/6) = 32/42 = 16/21