tough probability question

[demoskopico]

In a room 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 Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

1)5/21
2)3/7
3)4/7
4)5/7
5)16/21


someone gave this explanation: (4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.

20/42 + 12/42
32/42 = 16/21

E. 16/21

why do sum probability in this case? don't we need a multiplication in order for the two events to occur?

[kevin]

Can you see that there are two mutually exclusive cases

Either (1) the first person selected has only one friend or (2) the first person selected has two friends. You must find the probability of the desired result in each case and then sum

Remember for mutually exclusive A and B. Pr( A U B ) = Pr(A) + Pr(B).

Alternatively, you might like the following:

How many different subsets of two people can be formed from the seven people? 7C2 = 21

How many of these subsets contain two friends? (You try this)