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# In a room with 7 people, 4 people have exactly 1 friend in

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Intern
Joined: 13 Jun 2007
Posts: 48
In a room with 7 people, 4 people have exactly 1 friend in  [#permalink]

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08 Dec 2007, 12:00
5
4
00:00

Difficulty:

95% (hard)

Question Stats:

51% (02:36) correct 49% (02:02) wrong based on 210 sessions

### HideShow timer Statistics

In a room 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 Jane is Paul's friend, Paul is Jane'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

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
Director
Joined: 12 Jul 2007
Posts: 843

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08 Dec 2007, 12:09
4
2
(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.

20/42 + 12/42
32/42 = 16/21

E. 16/21
VP
Joined: 22 Nov 2007
Posts: 1040

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16 Jan 2008, 22:04
eschn3am wrote:
(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.

20/42 + 12/42
32/42 = 16/21

E. 16/21

excuse me, can you explain me this passage in detail? why do we have to sum probabilities up? don't we need to multiply them? we need that both conditions apply. thanks a lot
Director
Joined: 12 Jul 2007
Posts: 843

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17 Jan 2008, 04:51
1
We don't need to multiply them together because we're only choosing one pair of people.

$formdata=\frac{4}{7}*\frac{5}{6}$ = probability of choosing 2 people who aren't friends out of the 4 with one friend
$formdata=\frac{3}{7}*\frac{4}{6}$ = probability of choosing 2 people who aren't friends out of the 3 with two friends

Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.

7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way
VP
Joined: 22 Nov 2007
Posts: 1040

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18 Jan 2008, 04:17
eschn3am wrote:
We don't need to multiply them together because we're only choosing one pair of people.

$formdata=\frac{4}{7}*\frac{5}{6}$ = probability of choosing 2 people who aren't friends out of the 4 with one friend
$formdata=\frac{3}{7}*\frac{4}{6}$ = probability of choosing 2 people who aren't friends out of the 3 with two friends

Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.

7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way

right....let's solve it with combs:

How many different subsets of two people can be formed from the seven people? 7C2 = 21

How many of these subsets contain two friends? please give me a hand in doing this...
Director
Joined: 01 Jan 2008
Posts: 593

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18 Jan 2008, 07:00
6
marcodonzelli wrote:
right....let's solve it with combs:

How many different subsets of two people can be formed from the seven people? 7C2 = 21

How many of these subsets contain two friends? please give me a hand in doing this...

you are right that there are 21 possible ways of choosing a pair. I suggest visualizing a graph of friendships. 1 and 2 are friends; 3 and 4 are friends; 5, 6 and 7 are friends (3 pairs here: 5-6, 6-7, 5-7). Therefore, there are 5 total pairs of friends. Then probability of choosing a pair who are not friends is 1-5/21 = 16/21. I hope that helps.
CEO
Joined: 17 Nov 2007
Posts: 3439
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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19 Jan 2008, 03:27
1
an unusual way:

$$p=1-\frac{\frac{4*C_1^6+3*C_2^6}{2}}{C_2^7}=1-\frac{5}{21}=\frac{16}{21}$$
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Manager
Joined: 09 Apr 2008
Posts: 52

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13 Aug 2008, 23:30
walker wrote:
an unusual way:

$$p=1-\frac{\frac{4*C_1^6+3*C_2^6}{2}}{C_2^7}=1-\frac{5}{21}=\frac{16}{21}$$

Can you explain this a little more please? I understand the denominator - pick two people from 7. I think the numerator is the different combinations of people who ARE friends. But how did you get this?
CEO
Joined: 17 Nov 2007
Posts: 3439
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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14 Aug 2008, 11:13
1
$$4*C_1^6$$ - for each person out of 4 we have only one friend out of remained people (7-1=6)
$$3*C_2^6$$ - for each person out of 3 we have two friends out of remained people (7-1=6)

$$\frac{4*C_1^6+3*C_2^6}{2}$$ - excluding double counting.
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Senior Manager
Joined: 07 Jan 2008
Posts: 349

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19 Aug 2008, 22:54
my way:

$$p= 1-\frac{1+1+3}{C_2^7}= \frac{16}{21}$$
Manager
Joined: 27 Oct 2008
Posts: 177

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27 Sep 2009, 11:26
1
In a room 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 Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
1)5/21
2)3/7
3)4/7
4)5/7
5)16/21

Soln Assuming the people to be A,B,C,D,E,F,G

We can arrange in this way such that 3 people have two friends while 4 have 1 friend
A B
B A
C D E
D C E
E C D
F G
G F

First column represents the 7 people, second and third columns represent their friends.
Thus A,B,F,G each have one friend while C,D,E have two friends

Probability when two people are chosen where they are not friends is
= (5 + 5 + 2 + 2 + 2)/7C2
= 16/21

Ans is option 5
Intern
Joined: 09 Aug 2009
Posts: 46

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06 Oct 2009, 23:01
Only one friend
A<->B
C<->D
two friends
E<->F
E<->G
F<->G

so total possibility 21
=4*3+4/21
=16/21

good question........
Director
Joined: 17 Dec 2012
Posts: 625
Location: India
Re: In a room with 7 people, 4 people have exactly 1 friend in  [#permalink]

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23 Apr 2013, 05:47
alexperi wrote:
In a room 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 Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
1)5/21
2)3/7
3)4/7
4)5/7
5)16/21

1. We will find the reverse case, i.e., 2 people who are selected, are friends.
2. There are two groups here. People with 1 friend and people with 2 friends.
3. Friends can be selected as follows: (i) Among the 4 people each with just 1 friend, there will be 2 pairs of friends. (ii) Among the 3 people each with 2 friends, there will be 3 pairs of friends.
4. So 5 cases of friends can be selected.
5. The total number of ways of selecting 2 people is$$7C2 = 21$$
6. The number of cases when the 2 selected people are not friends $$= 21-5 = 16$$
7. Probability that the two slected people are not friends$$= 16/21$$
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Math Expert
Joined: 02 Sep 2009
Posts: 52161
Re: In a room with 7 people, 4 people have exactly 1 friend in  [#permalink]

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23 Apr 2013, 05:51
alexperi wrote:
In a room 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 Jane is Paul's friend, Paul is Jane'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

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
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Non-Human User
Joined: 09 Sep 2013
Posts: 9415
Re: In a room with 7 people, 4 people have exactly 1 friend in  [#permalink]

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31 Dec 2018, 21:38
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