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Picking up friends ....Permutation Combination

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Picking up friends ....Permutation Combination [#permalink]

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New post 19 Jan 2009, 23:38
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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?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

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Posts: 136

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Re: Picking up friends ....Permutation Combination [#permalink]

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New post 20 Jan 2009, 03:17
milind1979 wrote:
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?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

Group A: 4 people have exactly 1 friend in the room and Group B: 3 people have exactly 2 friends in the room

I pick 1 person. He could be member of Group A or B
The possibility of picking a member from Group A is: 4/7. The possibility of picking a person from 6 left that is not the first person's friend is 5/6.

The possibility of picking a member from Group B is 3/7. The possibility of picking another from 6 left that is not the first one's friend is 4/6

so the result IMO is:
4/7 * 5/6 + 3/7 * 4/6 = 16/21

Kudos [?]: 123 [0], given: 40

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Re: Picking up friends ....Permutation Combination [#permalink]

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New post 20 Jan 2009, 05:52
DavidArchuleta wrote:
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?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

Group A: 4 people have exactly 1 friend in the room and Group B: 3 people have exactly 2 friends in the room

I pick 1 person. He could be member of Group A or B
The possibility of picking a member from Group A is: 4/7. The possibility of picking a person from 6 left that is not the first person's friend is 5/6.

The possibility of picking a member from Group B is 3/7. The possibility of picking another from 6 left that is not the first one's friend is 4/6

so the result IMO is:
4/7 * 5/6 + 3/7 * 4/6 = 16/21
milind1979 wrote:
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?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

Group A: 4 people have exactly 1 friend in the room and Group B: 3 people have exactly 2 friends in the room

I pick 1 person. He could be member of Group A or B
The possibility of picking a member from Group A is: 4/7. The possibility of picking a person from 6 left that is not the first person's friend is 5/6.

The possibility of picking a member from Group B is 3/7. The possibility of picking another from 6 left that is not the first one's friend is 4/6

so the result IMO is:
4/7 * 5/6 + 3/7 * 4/6 = 16/21

[ Post details ]


Good job! DavidArchuleta
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Re: Picking up friends ....Permutation Combination [#permalink]

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New post 20 Jan 2009, 08:59
milind1979 wrote:
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?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21


A B C D --group 1 who have only one friend
E F G -- group 2 who have exactly two friends

try to find possible ways two selected are friends
AE
BE
CG
DF
FG
only 5 combinations possible

p= 1- 5/21 = 16/21
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Kudos [?]: 1032 [0], given: 5

Re: Picking up friends ....Permutation Combination   [#permalink] 20 Jan 2009, 08:59
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