In a room filled 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 John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
Solution: 2 people can be selected out of 7 in 7c2 = 21 ways
now 4 people have exactly one friend i.e. they will constitute 2 combinations e.g AB CD (mutually 4 people with 1 friend each)
& 3 with two friends will make 3 combinations i.e EF FH HE
total =5 combinations
so probability that both are friends is 5/21
so P (both are not friends ) =1-5/21 = 16/21
I DINT UNDERSTAND THE MUTUAL FRIENDSHIP THG AND THOSE 5 COMBINATIONS. PLZ ELABORATE AND EXPLAIN.
Solution: 2 people can be selected out of 7 in 7c2 = 21 ways
now 4 people have exactly one friend i.e. they will constitute 2 combinations e.g AB CD (mutually 4 people with 1 friend each)
& 3 with two friends will make 3 combinations i.e EF FH HE
total =5 combinations
so probability that both are friends is 5/21
so P (both are not friends ) =1-5/21 = 16/21
I DINT UNDERSTAND THE MUTUAL FRIENDSHIP THG AND THOSE 5 COMBINATIONS. PLZ ELABORATE AND EXPLAIN.
