can u post the answer.....if possible the steps.....i am stuck in this one.

Begin by counting the number of relationships that exist among the 7 individuals whom we will call A, B, C, D, E, F, and G.

First consider the relationships of individual A: AB, AC, AD, AE, AF, AG = 6 total. Then consider the relationships of individual B without counting the relationship AB that was already counted before: BC, BD, BE, BF, BG = 5 total. Continuing this pattern, we can see that C will add an additional 4 relationships, D will add an additional 3 relationships, E will add an additional 2 relationships, and F will add 1 additional relationship. Thus, there are a total of 6 + 5 + 4 + 3 + 2 + 1 = 21 total relationships between the 7 individuals. Alternatively, this can be computed formulaically as choosing a group of 2 from 7: 7!/(2! 5!) = 21

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

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.

To get to the answer in 2-2.5 minute range is certainly not easy. I had to think about this one for about 3-4 minutes before I tried to answer it. Even then that took me way too long. Maybe someone out there has a shortcut for solving these types of questions.

Cheers
K

The way Kaunteya has done it is the best and the only way to arrive at an answer for such questions.
If we try to do it any faster than this, we are likely to make a mistake while solving.
Such problems are typical of Manhattan tests, but its unlikely that such questions will occur in GMAT test unless we've been scoring very highly (getting a lot of questions in 700+ difficulty level right)..
To make it faster we can try to eliminate obviously wrong options and guess among the likely ones.. Can't think of any other way.

p=1-(4/7*1/6+3/7*2/6)=16/21

4/7 - the probability to choose one of 4 people who have 1 friend.
1/6 - the probability to choose a friend of 6 remained people

3/7 - the probability to choose one of 3 people who have 2 friend.
2/6 - the probability to choose a friend of 6 remained people.