In a group of 7 people, 4 have exactly one sibling and 3 have exactly two siblings. Two people selected at random from the group, what is the probability that they are NOT siblings?

Concept:

Number of ways of selecting r people from n people is \(^nC_r\)

Explanation:

Number of ways of selecting 2 people randomly from 7 people is \(^7C_2=\frac{7\times 6}{2}=21\)

Now, Let 4 peoples are A, B, C, D

They have  exactly one sibling so the pair of siblings can be (A, B), (C, D)

Also let 3 people be X, Y, Z

They have exactly two siblings so the pair of siblings can be (X, Y), (X, Z), (Y, Z)

So favourable outcomes are (A, C), (A, D) (B, C), (B, D),(A, X), (A, Y), (A, Z), (B, X), (B, Y), (B, Z), (C, X), (C, Y), (C, Z), (D, Y) (D, X), (D, Z)

Therefore probability that they are siblings = \(\frac{16}{21}\) 

Option (2) is correct