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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?

sorry but im still trying to figure out the arrangement. let A, B, C, D, E, F, and G be the 7 guys.

4 people exactly having 1 friend wud mean AB, AC, AD,AE. that makes B,C,D,E have just A as friend.(is this assumption correct?)

But, to make it simple, the question gives the following assumption (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend).

sorry but im still trying to figure out the arrangement. let A, B, C, D, E, F, and G be the 7 guys.

4 people exactly having 1 friend wud mean AB, AC, AD,AE. that makes B,C,D,E have just A as friend.(is this assumption correct?)

But, to make it simple, the question gives the following assumption (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend).

So does it mean that in the group of 4 people having only 1 friend, can i assume that we just need 4 people to make up this group (instead of 5) because if A is a friend of B, and B is also a friend of A. I am very confused with this type of question.

5/6 instead of 3/6, coz there are 2 others in the same group(4) who'd be fair game too. They are friends with each other and not necessarily friends with our possible pick. Hence you have to consider picking them as favorable outcome(which totals 3 in group(3) and 2 from group(4)). Ofcourse total # of outcomes would be 6(7-1). HTH...
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I started the problem trying to find out what is the probability of two people picked are friends

Total combo's = 7 C 2 = 21

{(a,b) (c,d)} (e,f,g) are such groups that fit the description

2 people can be picked from any of the 3 groups in 3 ways.

2 people out of (a,b) can be picked in 1 way. Similarly in 1 way out of (c,d). 3 c 2 ways from (e,f,g)

I got 9/21 ways for them being friends. How ever, the answer seems to be 5/21 as the correct answer for not being friends is 16/21. Where did I go wrong fellas?

5/6 instead of 3/6, coz there are 2 others in the same group(4) who'd be fair game too. They are friends with each other and not necessarily friends with our possible pick. Hence you have to consider picking them as favorable outcome(which totals 3 in group(3) and 2 from group(4)). Ofcourse total # of outcomes would be 6(7-1). HTH...

I started the problem trying to find out what is the probability of two people picked are friends

Total combo's = 7 C 2 = 21

{(a,b) (c,d)} (e,f,g) are such groups that fit the description

2 people can be picked from any of the 3 groups in 3 ways.

2 people out of (a,b) can be picked in 1 way. Similarly in 1 way out of (c,d). 3 c 2 ways from (e,f,g)

I got 9/21 ways for them being friends. How ever, the answer seems to be 5/21 as the correct answer for not being friends is 16/21. Where did I go wrong fellas?

hi icandy, I think with your way of thinking, you have 1 way of picking 2 people from (a,b) 1 way of picking 2 people from (c,d) 3 way of picking 2 people from (e,f,g) (3!/2!) Total will be 5 ways You have a total combo is 21 So the Probability of picking 2 people who are friends is 5/21. So the probability of picking 2 people who are not friends is 16/21. Is that correct?

I started the problem trying to find out what is the probability of two people picked are friends

Total combo's = 7 C 2 = 21

{(a,b) (c,d)} (e,f,g) are such groups that fit the description

2 people can be picked from any of the 3 groups in 3 ways.

2 people out of (a,b) can be picked in 1 way. Similarly in 1 way out of (c,d). 3 c 2 ways from (e,f,g)

I got 9/21 ways for them being friends. How ever, the answer seems to be 5/21 as the correct answer for not being friends is 16/21. Where did I go wrong fellas?

YOU MISSED THESE TWO FRIENDS: (a,b) (c,d).

So total friends = 5

so the desired prob = 1 - 5/21 = 16/21
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