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In a room filled with 7 people, 4 people have exactly 1

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In a room filled with 7 people, 4 people have exactly 1 [#permalink]  14 Jul 2008, 18:24
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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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Re: PS - probability that those two individuals are not friends? [#permalink]  14 Jul 2008, 18:52
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

lets see total possibility = 7C2 =21

so out of the 4,.we have 4c1 ways to pick 1 from this group of 4
and 3c1 from out of the other group 3c1..

4c1*3c1=12/27 4/7
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Re: PS - probability that those two individuals are not friends? [#permalink]  14 Jul 2008, 19:06
x-ALI-x 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

Possible combinatios of selecting 2 people from 7 = 7C2 = 21

Possible Combinations of finding 2 people i.e a pair such that both are friends = 5
(I calculated this following way, 7 individuals named 1,2,..7;
4 have exactly one friend: 1-2, 3-4
3 have exactly 2 friends: 5-6, 5-7, 6-7
)

Probability of getting 2 people who are friends = 5/21

Probabilty of getting 2 people who are not friends =
=1 - Probability of getting 2 people who are friends
= 1- 5/21
= 16/21

Thus E
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Re: PS - probability that those two individuals are not friends? [#permalink]  14 Jul 2008, 20:41
the question is quite simple the 4 guys who ONLY know 1 person other than themselves is one group of 4 people..then out of the 7 only the other 3 know each other...but these 2 groups dont know anyone from outside of their group..

so the question is asking pick one guy from 1 group (i.e the 4 guys who only have 1 friend) and one guy from group 2 (i.e the group of 3 socialites)..

so basically you have 4c1 and 3c1 possibilites picking 2 people..

total possibility of picking any 2 people is 7c2=27..

alpha_plus_gamma wrote:
x-ALI-x 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

Possible combinatios of selecting 2 people from 7 = 7C2 = 21

Possible Combinations of finding 2 people i.e a pair such that both are friends = 5
(I calculated this following way, 7 individuals named 1,2,..7;
4 have exactly one friend: 1-2, 3-4
3 have exactly 2 friends: 5-6, 5-7, 6-7
)

Probability of getting 2 people who are friends = 5/21

Probabilty of getting 2 people who are not friends =
=1 - Probability of getting 2 people who are friends
= 1- 5/21
= 16/21

Thus E
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Re: PS - probability that those two individuals are not friends? [#permalink]  14 Jul 2008, 21:10
fresinha12 wrote:
the question is quite simple the 4 guys who ONLY know 1 person other than themselves is one group of 4 people..then out of the 7 only the other 3 know each other...but these 2 groups dont know anyone from outside of their group..

so the question is asking pick one guy from 1 group (i.e the 4 guys who only have 1 friend) and one guy from group 2 (i.e the group of 3 socialites)..

so basically you have 4c1 and 3c1 possibilites picking 2 people..

total possibility of picking any 2 people is 7c2=27..

you can pick two guys from group 1 and still they wont be fried with eachother
1234 is the group of 4 guys, 1 is fried with 2 and 3 is frid with 4.... this doesnt mean that 1 is friend with 3

so you can select (1-3, 1-4, 2-3, 2-4) from the same group...... so you have to add 4 to 12 for total fav

Re: PS - probability that those two individuals are not friends?   [#permalink] 14 Jul 2008, 21:10
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