GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 26 Mar 2019, 20:07

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Director
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 925
Location: India
GMAT 1: 410 Q35 V11
GMAT 2: 530 Q44 V20
GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)
Re: In a room filled with 7 people, 4 people  [#permalink]

### Show Tags

03 Nov 2012, 17:36
1
Bunuel wrote:
Archit143 wrote:
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?

5/21
3/7
4/7
5/7
16/21

Merging similar topics. Please refer to the solutions on page 1 (foe example: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html#p645861)

Hope it helps.

BUNEL

Here is how i approached the problem and got the wrong answer can you pls correct me.

Total outcome = 7C2 = 21

Prob of getting exactly 2 sibbling = [4C1 * 3C1] + 3C2 i.e. [ 1 from 4 * 1 from 3] + [ 2 out of 3, since the group of 3 has 2 sibbling]
= [4*3] + 3 =15

prob of getting sibbling = 15 /21

prob of not getting sibbling = 1 - 15/21 = 6/21
Senior Manager
Joined: 13 Aug 2012
Posts: 419
Concentration: Marketing, Finance
GPA: 3.23
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

12 Nov 2012, 05:38
1
1
Ways to select the those 4 with their sibling?

4/7 x 1/6 = 4/42

Ways to select those 3 with one of their 2 siblings?

3/7 x 2/6 = 6/42

P = 1 - (4/42 + 6/42) = 1 - 10/42 = (42 -10)/42 = 32/42 = 16/21

P = 16/21

_________________
Impossible is nothing to God.
Intern
Joined: 11 Dec 2012
Posts: 2
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

13 Dec 2012, 15:08
well, this is how i approached it...

Total 7 people - 1,2,3,4,5,6,7
4 people have 1 sibling - [1,2];[3,4] - 2 single-sibling groups
3 people have 2 siblings - [5,6,7] - 1 two-siblings group

Total ways to select 2 people, 7C2 = 21.
Ways to select only siblings : 2C1( ways to select 1 group from 2 single-sibling groups) + 3C2( ways to select 2 people from 2 sibling-group) = 2+3= 5

Probability that NO siblings selected = 1- 5/21 = 16/21.

Hence, D.

Please let me know if my approach is flawed.
_________________
What happens with me is less significant than what happens within me!
Consider giving Kudos, appreciation is divine.
Intern
Joined: 04 Jan 2013
Posts: 13
Location: India
Concentration: Finance
GMAT Date: 08-26-2013
GPA: 2.83
WE: Other (Other)
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

15 May 2013, 01:26
WarriorGmat wrote:
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

If A is a sibling of B, then B is also a sibling of A.
If A is a sibling of B, and B is a sibling of C, that means A is also a sibling of C.

For first four people to have exactly one sibling each means two pairs of siblings.
For last 3 people to have exactly two siblings each means one triplet of siblings.

Now, to select two individuals from a group of 7, total no. of ways it can be done = 7C2 = 21

Cases if siblings are selected : 1. Pair 1
2. Pair 2
3,4,5 = 2 individuals from triplet of siblings = 3C2 = 3 ways.

So, there are 5 cases in which selected individuals are siblings.

Therefore, probability of two individuals selected are NOT sibligs is (21-5)/21 = 16/21
Intern
Joined: 09 Feb 2013
Posts: 7
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

15 May 2013, 01:33
mkdureja wrote:
WarriorGmat wrote:
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

If A is a sibling of B, then B is also a sibling of A.
If A is a sibling of B, and B is a sibling of C, that means A is also a sibling of C.

For first four people to have exactly one sibling each means two pairs of siblings.
For last 3 people to have exactly two siblings each means one triplet of siblings.

Now, to select two individuals from a group of 7, total no. of ways it can be done = 7C2 = 21

Cases if siblings are selected : 1. Pair 1
2. Pair 2
3,4,5 = 2 individuals from triplet of siblings = 3C2 = 3 ways.

So, there are 5 cases in which selected individuals are siblings.

Therefore, probability of two individuals selected are NOT sibligs is (21-5)/21 = 16/21

Hi mkdureja,
Thanks.
Intern
Joined: 04 Jan 2013
Posts: 13
Location: India
Concentration: Finance
GMAT Date: 08-26-2013
GPA: 2.83
WE: Other (Other)
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

15 May 2013, 02:01
1
Hi,

Que:
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?

Case 1:
4 people have exactly 1 sibling :
Let A, B, C, D be those 4 people
If A is sibling of B, then B is also a sibling of A.
Basically, they have to exist in pairs
Therefore, these four people compose of two pairs of siblings.

Case 2:
3 people have exactly 2 siblings :
This must obviously be a triplet of siblings.
Let them be E, F, G.
If E is sibling of F, and also is sibling of G., it means F and G are also siblings of each other.
E is sibling of exactly 2 : F and G
F is sibling of exactly 2 : E and G
G is sibling of exactly 2 : E and F
This group has three pairs of siblings.

Now, selecting two individuals out of the group of 7 people has:
1) Two Pairs, as in Case 1.
2) Three Pairs, as in Case 2.
So, 5 cases out of a total of 7C2 = 21 cases.

Hope it clarifies. If you have any further doubt, please point to exactly where you are having a problem understanding it.
Math Expert
Joined: 02 Sep 2009
Posts: 53865
In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

05 Jul 2013, 02:26
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

Similar question to practice: a-dog-breeder-currently-has-9-breeding-dogs-6-of-the-dogs-131992.html

*New project from GMAT Club!!! Check HERE

_________________
Director
Joined: 17 Dec 2012
Posts: 626
Location: India
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

05 Jul 2013, 22:07
2
1
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

It is first important to understand the problem. So let us first assume a specific case.

1. Assume the 7 people are A,B, C, D, E, F and G
2. Assume the 4 people who have 1 sibling are A, B, C and D
3. Let's assume A's sibling is B. Therefore B's sibling is A. Similarly for C and D.
4. So we are left with E, F and G. Each should have exactly 2 siblings.
5. E's siblings will be F and G. So F's siblings will be E and G and G's siblings will be E and F
6. Now look at the general case.
7.Total number of ways of selecting 2 people out of 7 people is 7C2=21
8. Instead of A, B, C and D assume any 4 people. We can see for every such 4 people assumed, there are 2 cases where the selected 2 will be siblings. In the case we assumed they are A and B or C and D. This gives one of the favorable outcomes
9. Or the 2 people selected being siblings may come out of the 3 siblings. The number of favorable outcomes is 3 as we can see in the specific case they are E and F, or F and G or E and G.
10. The total number of favorable outcomes for the selected two being siblings is 2+3=5.
11. The probability that the two selected are siblings is 5/21.
12, Therefore the probability that the two selected are not siblings is 1-5/21= 16/21
_________________
Srinivasan Vaidyaraman
Sravna Holistic Solutions
http://www.sravnatestprep.com

Holistic and Systematic Approach
Intern
Joined: 10 Dec 2013
Posts: 18
Location: India
Concentration: Technology, Strategy
Schools: ISB '16 (S)
GMAT 1: 710 Q48 V38
GPA: 3.9
WE: Consulting (Consulting)
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

14 Jan 2014, 21:53
Quote:
Sure.

We have the following siblings: {1, 2}, {3, 4} and {5, 6, 7}.

Now, in order to select two individuals who are NOT siblings we must select EITHER one from {5, 6, 7} and ANY from {1, 2} or {3, 4} OR one from {1, 2} and another from {3, 4}.

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

3/7 - selecting a sibling from {5, 6, 7}, 4/6 - selecting any from {1, 2} or {3, 4}. Multiplying by 2 since this selection can be don in two ways: the first from {5, 6, 7} and the second from {1, 2} or {3, 4} OR the first from {1, 2} or {3, 4} and the second from {5, 6, 7};

2/7 - selecting a sibling from {1, 2}, 2/6 - selecting a sibling from {3, 4}. Multiplying by 2 since this selection can be don in two ways: the first from {1, 2} and the second from {3, 4} OR the first from {3, 4} and the second from {1, 2}.

Other approaches here: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html#p645861

Hope it's clear.

Hi Bunuel,
Can you please explain to me that why when we use simple probability we need to consider both the cases of selection, if we pick a sibling from the first group of 2 siblings first and then we pick a sibling from the second group of 2 siblings and if we pick a sibling from the second group of 2 siblings first and then we pick a sibling from the first group of 2 siblings.
While if we use combinomatric approach we just count one case i.e. 2C1*2C1 which is for selecting one each from the 2 different 2 sibling groups.
Manager
Joined: 19 Mar 2012
Posts: 98
Location: United States
Schools: IIMA PGPX"20

### Show Tags

26 Jan 2014, 11:37
Hi Bunuel
Please correct me where I went wrong.
I assumed the 7 siblings in this way.PFA the image.
So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each.
Now to select two people that are not siblings :
1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5)
2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4)
So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

Attachments

File comment: My Solution

Questions.jpg [ 20.61 KiB | Viewed 6681 times ]

_________________
Feel Free to Press Kudos if you like the way I think .
Math Expert
Joined: 02 Sep 2009
Posts: 53865

### Show Tags

27 Jan 2014, 01:28
282552 wrote:
Hi Bunuel
Please correct me where I went wrong.
I assumed the 7 siblings in this way.PFA the image.
So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each.
Now to select two people that are not siblings :
1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5)
2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4)
So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.
_________________
Manager
Joined: 19 Mar 2012
Posts: 98
Location: United States
Schools: IIMA PGPX"20

### Show Tags

27 Jan 2014, 06:33
Bunuel wrote:
282552 wrote:
Hi Bunuel
Please correct me where I went wrong.
I assumed the 7 siblings in this way.PFA the image.
So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each.
Now to select two people that are not siblings :
1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5)
2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4)
So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.

I still do not understand .The question simply says "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". and my diagram meets the given criterion.Also is it necessary that if A&B and B& C are siblings then A&C also need to be siblings.
_________________
Feel Free to Press Kudos if you like the way I think .
Math Expert
Joined: 02 Sep 2009
Posts: 53865

### Show Tags

27 Jan 2014, 06:36
282552 wrote:
Bunuel wrote:
282552 wrote:
Hi Bunuel
Please correct me where I went wrong.
I assumed the 7 siblings in this way.PFA the image.
So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each.
Now to select two people that are not siblings :
1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5)
2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4)
So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.

I still do not understand .The question simply says "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". and my diagram meets the given criterion.Also is it necessary that if A&B and B& C are siblings then A&C also need to be siblings.

Do you know what a sibling mean? How can (A, B), (B, C), (C, D) be BROTHERS, and (A, C), (A, D), and (B, D) not to be?
_________________
Intern
Joined: 10 Dec 2013
Posts: 18
Location: India
Concentration: Technology, Strategy
Schools: ISB '16 (S)
GMAT 1: 710 Q48 V38
GPA: 3.9
WE: Consulting (Consulting)
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

29 Jan 2014, 11:33
Hi,

Can someone please explain to me that why when we use simple probability we need to consider both the cases of selection i.e. if we pick a sibling from the first group of 2 siblings(A,B) first and then we pick a sibling from the second group of 2 siblings(C,D) and if we pick a sibling from the second group of 2 siblings(C,D) first and then we pick a sibling from the first group of 2 siblings(A,B)
While if we use combinomatric approach we just count one case i.e. 2C1*2C1 which is for selecting one each from the 2 different 2 sibling groups

How is A,C and C,A different if we just have to check that whether the two selected are siblings or not? And if it makes a difference why we did not you permutation instead of combinations?
Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 757
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

19 May 2014, 00:49
$$2*\frac{1}{7}*\frac{5}{6}+2*\frac{1}{7}*\frac{5}{6}+3*\frac{1}{7}*\frac{4}{6}$$
$$\frac{32}{42} = \frac{16}{21}$$
_________________
Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos
My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
Intern
Joined: 14 Feb 2014
Posts: 6
GMAT 1: 550 Q45 V21
GMAT 2: 690 Q49 V35
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

08 Oct 2014, 04:16
Attachments

probability.JPG [ 40.76 KiB | Viewed 5883 times ]

Math Expert
Joined: 02 Sep 2009
Posts: 53865
In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

08 Oct 2014, 04:43
1
keyrun wrote:

When you select groups 1 and 2, you have 3*2 choices, when groups 1 and 3 you have 3*2 choices and when groups 2 and 3 you have 2*2 choices: total = 6 + 6 + 4.
_________________
Manager
Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 203
Location: India
MISSION : 800
WE: Design (Manufacturing)
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

30 Jun 2015, 00:43
Hi everyone

I was stuck with this question b/w the no of pairs of siblings and the answer

I thought i should post this.

May be it helps someone out dr.

Happy learning
Attachments

sib.jpg [ 149.93 KiB | Viewed 5113 times ]

_________________
Thank you

+KUDOS

> I CAN, I WILL <
Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 414
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

08 Jul 2015, 10:00
I am posting a visual solution.

To begin with, there are 7!/2!*5! = 21 total possibile pairings.

After finding the no-sibling pairings (see image below) we end up with 16/21.
Attachments

siblings.png [ 26.7 KiB | Viewed 4134 times ]

Manager
Joined: 10 Mar 2013
Posts: 189
GMAT 1: 620 Q44 V31
GMAT 2: 690 Q47 V37
GMAT 3: 610 Q47 V28
GMAT 4: 700 Q50 V34
GMAT 5: 700 Q49 V36
GMAT 6: 690 Q48 V35
GMAT 7: 750 Q49 V42
GMAT 8: 730 Q50 V39
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

### Show Tags

10 Jul 2015, 20:46
x = P(1-sibling person) = 4*p(1-sibling person)
y = P(2-sibling person) = 3*p(2-sibling person)

p(1-sibling person)=1/7*1/6 (1/6 because we have 1 sibling in the 6 remaining people)
p(2-sibling person)=1/7*2/6 (2/6 because we have 2 siblings in the 6 remaining people)

E
Re: In a room filled with 7 people, 4 people have exactly 1   [#permalink] 10 Jul 2015, 20:46

Go to page   Previous    1   2   3    Next  [ 48 posts ]

Display posts from previous: Sort by