In a room filled with 7 people, 4 people have exactly 1 sibl : GMAT Problem Solving (PS)
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# In a room filled with 7 people, 4 people have exactly 1 sibl

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In a room filled with 7 people, 4 people have exactly 1 sibl [#permalink]

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14 Jul 2008, 17:24
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In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
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Re: In a room filled with 7 people, 4 people have exactly 1 sibl [#permalink]

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16 Sep 2013, 00:23
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Expert's post
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4).
As there are 3 people with exactly 2 siblings each: we have one triple of siblings (4-5-6).

Solution #1:
# of selections of 2 out of 7 - $$C^2_7=21$$;
# of selections of 2 people which are not siblings - $$C^1_2*C^1_2$$ (one from first pair of siblings*one from second pair of siblings)+$$C^1_2*C^1_3$$ (one from first pair of siblings*one from triple)+ $$C1^_2*C^1_3$$(one from second pair of siblings*one from triple) $$=4+6+6=16$$.

$$P=\frac{16}{21}$$

Solution #2:
# of selections of 2 out of 7 - $$C^2_7=21$$;
# of selections of 2 siblings - $$C^2_3+C^2_2+C^2_2=3+1+1=5$$;

$$P=1-\frac{5}{21}=\frac{16}{21}$$.

Solution #3:
$$P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
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Re: PS - probability that those two individuals are not friends? [#permalink]

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14 Jul 2008, 18:06
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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: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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16 Sep 2013, 00:16
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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.

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.
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Re: PS - probability that those two individuals are not friends? [#permalink]

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14 Jul 2008, 17:52
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]

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14 Jul 2008, 19:41
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]

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14 Jul 2008, 20:10
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, 20:10
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