Probability: Tricky qn.

[Faith 26]

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?
- 5/21
- 3/7
- 4/7
- 5/7
- 16/21

[Faith 26]

Source: HelloChampion.com

[EDITED BY STACEY: This is actually one of ours.]

[Guest]

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?
- 5/21
- 3/7
- 4/7
- 5/7
- 16/21

Is the OA 4/7? I 'll let u know of my approach if my answer is correct

[Harish Dorai]

I think it is 16/21. My approach was follows.

We can pick 2 persons as follows:

Situation 1) Person 1 selected is the one with just one friend and Person 2 is not his friend.
Situation 2) Person 2 selected is the one with two friends and Person 2 is not his friend.

so the probability is sum of the probabilities of above situations.

For situation 1 it is 4/7 multiplied by 5/6 (Probability to pick a person with 1 friend is 4/7, and out of the remaining 6, 5 are not his friends).

For situation 2 it is 3/7 multiplied by 4/6 (Probability to pick a person with 2 friends is 3/7 and out of the remaining 6, 4 are not his friends).

So probability is = 4/7 x 5/6 + 3/7 x 4/6 which when simplified becomes 16/21.

[givemeanid]

2/7*5/6 + 2/7*5/6 + 3/7*4/6 = 10/42 + 10/42 + 12/42 = 32/42 = 16/21

[anadi]

Total Number of ways to select 2 people out of 7, 7!/5!2! = 21
Based on the conditions given, if a,b,c,d,e,f,g are the people in the room, and b are friends, c and d are friends, and e,f,and g are friends of one another. So total number of ways to select 2 people WHO ARE FRIENDS is, 1 (a and b) +1 (c and d) +3 (2 from e,f, and g) = 5.

Hence total number of selecting people who are not friends is 21-5 = 16. So probability is 16/21.

[StaceyKoprince]

Official explanation:

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.