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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

09 Jul 2009, 19:39

1

This post received KUDOS

14

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

49% (02:14) correct
51% (00:55) wrong based on 314 sessions

HideShow timer Statistics

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?

Re: Probability - Siblings in the room [#permalink]

Show Tags

10 Jul 2009, 03:47

10

This post received KUDOS

2

This post was BOOKMARKED

I3igDmsu 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

I divide 7 people into 2 groups: Group 1 including 4 people have exactly 1 sibling and Group 2 including 3 people have exactly 2 siblings First person: If he is from G1 (probability: 4/7), the probability that the second one is not his sibling: 5/6 so 4/7*5/6=10/21 If he is from G2 (probability: 3/7), the probability that the second one is not his sibling: 4/6 so 3/7*4/6=6/21 10/21+6/21=16/21

Re: Probability - Siblings in the room [#permalink]

Show Tags

10 Jul 2009, 09:26

1

This post was BOOKMARKED

I have a basic question: If A is a sibling of B and B is a sibling of C, do we consider A as a sibling of C? Then there should be more than 7 ppl in the room.

If we don't assume the above, then I get 4/7. We have two mutually exclusive groups. A grp of 4 where each has one sibling, forms a square with vertices as ppl and edges as relationship.

A grp of 3 where each has two siblings, forms a triangle with vertices as ppl and edges as relationship.

Basically, we have to get one guy from G1 and other from G2. So , I get, 4C1 * 3C1 / 7C2 = 4/7

Any comments? explanations?

DavidArchuleta wrote:

I3igDmsu 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

I divide 7 people into 2 groups: Group 1 including 4 people have exactly 1 sibling and Group 2 including 3 people have exactly 2 siblings First person: If he is from G1 (probability: 4/7), the probability that the second one is not his sibling: 5/6 so 4/7*5/6=10/21 If he is from G2 (probability: 3/7), the probability that the second one is not his sibling: 4/6 so 3/7*4/6=6/21 10/21+6/21=16/21

Re: Probability - Siblings in the room [#permalink]

Show Tags

10 Jul 2009, 09:48

14

This post received KUDOS

6

This post was BOOKMARKED

If 4 people have exactly 1 sibling, there are 2 pairs of siblings in the 4 people

Say A B C and D, each have exactly 1 sibling. That implies, A has 1 sibling (say B), so B's sibling is A Similarly, C and D are siblings

Now if 3 people (say E F and G) each have 2 siblings, then E must have F and G both as siblings F must have E and G both as siblings, and same for G too.

So we have 2 pairs of 2 siblings and 1 triplet of 3 siblings A-B C-D E-F-G

ways of selecting 2 out of 7 is = 21 ways ways of NOT selecting a sibling pair = Total ways - ways of always getting a pair

ways of getting a sibling pair : A-B = 1way C-D = 1 way 2 out of 3 (E - F - G) = 3 ways

so 5 ways we always get a sibling pair, remaining 21-5 ways we dont get a sibling pair

Re: Probability - Siblings in the room [#permalink]

Show Tags

10 Jul 2009, 10:24

1

This post received KUDOS

awesome...+1

arghhh..i was only thinking about one way relationship..A-B = B-A and hence we have two pairs..true

rashminet84 wrote:

If 4 people have exactly 1 sibling, there are 2 pairs of siblings in the 4 people

Say A B C and D, each have exactly 1 sibling. That implies, A has 1 sibling (say B), so B's sibling is A Similarly, C and D are siblings

Now if 3 people (say E F and G) each have 2 siblings, then E must have F and G both as siblings F must have E and G both as siblings, and same for G too.

So we have 2 pairs of 2 siblings and 1 triplet of 3 siblings A-B C-D E-F-G

ways of selecting 2 out of 7 is = 21 ways ways of NOT selecting a sibling pair = Total ways - ways of always getting a pair

ways of getting a sibling pair : A-B = 1way C-D = 1 way 2 out of 3 (E - F - G) = 3 ways

so 5 ways we always get a sibling pair, remaining 21-5 ways we dont get a sibling pair

Re: Probability - Siblings in the room [#permalink]

Show Tags

16 May 2011, 05:21

Im missing something.

groups are a-b c-d e-f-g

total options 21

if i choose A - i can choose c-d-e-f-g (5 options) if i choose B - i can choose c-d-e-f-g (5 options) if i choose C - i can choose a-b-e-f-g (5 options) if i choose D - i can choose a-b-e-f-g (5 options)

and more. if i choose one from the triple group.so what am i missing? _________________

Re: Probability - Siblings in the room [#permalink]

Show Tags

16 May 2011, 05:42

144144 wrote:

Im missing something.

groups are a-b c-d e-f-g

total options 21

if i choose A - i can choose c-d-e-f-g (5 options) if i choose B - i can choose c-d-e-f-g (5 options) if i choose C - i can choose a-b-e-f-g (5 options) if i choose D - i can choose a-b-e-f-g (5 options)

and more. if i choose one from the triple group.so what am i missing?

Nothing. Keep going the way you started. You will end up with 16 favorable cases.

P=16/21

OR

Method1(Combination):

Group1: a,b Group2: c,d Group3: e,f,g

Choose 1 from Group1 AND Choose 1 from Group2 OR Choose 1 from Group2 AND Choose 1 from Group3 OR Choose 1 from Group1 AND Choose 1 from Group3

We know AND=* and OR=+

Favorable cases: C(2,1)*C(2,1)+C(2,1)*C(3,1)+C(2,1)*C(3,1)=2*2+2*3+2*3=4+6+6=16 Total C(7,2)=21 P=16/21

Re: Probability - Siblings in the room [#permalink]

Show Tags

16 May 2011, 10:07

fluke wrote:

144144 wrote:

Im missing something.

groups are a-b c-d e-f-g

total options 21

if i choose A - i can choose c-d-e-f-g (5 options) if i choose B - i can choose c-d-e-f-g (5 options) if i choose C - i can choose a-b-e-f-g (5 options) if i choose D - i can choose a-b-e-f-g (5 options)

and more. if i choose one from the triple group.so what am i missing?

Nothing. Keep going the way you started. You will end up with 16 favorable cases.

P=16/21

OR

Method1(Combination):

Group1: a,b Group2: c,d Group3: e,f,g

Choose 1 from Group1 AND Choose 1 from Group2 OR Choose 1 from Group2 AND Choose 1 from Group3 OR Choose 1 from Group1 AND Choose 1 from Group3

We know AND=* and OR=+

Favorable cases: C(2,1)*C(2,1)+C(2,1)*C(3,1)+C(2,1)*C(3,1)=2*2+2*3+2*3=4+6+6=16 Total C(7,2)=21 P=16/21

Im already at 20 possibilities without even the 3rd group. 5+5+5+5 (=20) out of 21 possible options. so its not working. I am missing something here. _________________

Re: Probability - Siblings in the room [#permalink]

Show Tags

16 May 2011, 18:25

11

This post received KUDOS

Expert's post

Alchemist1320 wrote:

can we ever expect so tough questions in GMAT ?

wondering if I can ever get it

Anyways thanks for the all the replies...helps in devloping line of thinking

The question is not very difficult. Try to take one line at a time to reason it out. I will tell you what I mean. I read the first line:

"In a room filled with 7 people, 4 people have exactly 1 sibling in the room and"

I stop here. 4 people have exactly 1 sibling. So I say to myself, "Ok. A is there and he has a sibling B. Wait, this means that automatically B has the sibling A. They cannot have any other sibling since both should have only one sibling each. Then out of 4 people, 2 are already accounted for. Then in the same way, there must be C there and his sibling D". The important thing to realize is that 'sibling' relation is symmetric. If A is sibling of B, B is also the sibling of A.

Next line of the question: "3 people have exactly 2 siblings in the room." Now I think, "There is E there and he has two siblings F and G. So automatically F and G both have 2 siblings each too."

This gives me following siblings: A - B C - D E - F - G

Now I need to pick 2 people such that they are not siblings. I can do it in 2 ways. I can either find the number of ways of picking siblings or number of ways of picking 'non siblings'. Number of ways of picking siblings is certainly easier - I can pick A-B or C-D or 2 of E-F-G in 3 ways (E-F or F-G or E-G) so there are a total of 5 ways of picking siblings. Total ways of picking 2 people is 7C2 = 21 ways Out of these 21 ways, 5 ways are there to pick 2 siblings so other 16 ways must be of picking 'non siblings'.

Re: Probability - Siblings in the room [#permalink]

Show Tags

21 Jun 2011, 22:38

VeritasPrepKarishma wrote:

Alchemist1320 wrote:

can we ever expect so tough questions in GMAT ?

wondering if I can ever get it

Anyways thanks for the all the replies...helps in devloping line of thinking

The question is not very difficult. Try to take one line at a time to reason it out. I will tell you what I mean. I read the first line:

"In a room filled with 7 people, 4 people have exactly 1 sibling in the room and"

I stop here. 4 people have exactly 1 sibling. So I say to myself, "Ok. A is there and he has a sibling B. Wait, this means that automatically B has the sibling A. They cannot have any other sibling since both should have only one sibling each. Then out of 4 people, 2 are already accounted for. Then in the same way, there must be C there and his sibling D". The important thing to realize is that 'sibling' relation is symmetric. If A is sibling of B, B is also the sibling of A.

Next line of the question: "3 people have exactly 2 siblings in the room." Now I think, "There is E there and he has two siblings F and G. So automatically F and G both have 2 siblings each too."

This gives me following siblings: A - B C - D E - F - G

Now I need to pick 2 people such that they are not siblings. I can do it in 2 ways. I can either find the number of ways of picking siblings or number of ways of picking 'non siblings'. Number of ways of picking siblings is certainly easier - I can pick A-B or C-D or 2 of E-F-G in 3 ways (E-F or F-G or E-G) so there are a total of 5 ways of picking siblings. Total ways of picking 2 people is 7C2 = 21 ways Out of these 21 ways, 5 ways are there to pick 2 siblings so other 16 ways must be of picking 'non siblings'.

Required probability = 16/21

Can someone please explain this ?

I can pick A-B or C-D or 2 of E-F-G in 3 ways (E-F or F-G or E-G) so there are a total of 5 ways of picking siblings. Total ways of picking 2 people is 7C2 = 21 ways Out of these 21 ways, 5 ways are there to pick 2 siblings so other 16 ways must be of picking 'non siblings'.

Required probability = 16/21[/quote]

And also, How do we get 7C2?? Are there 7 possible cases?

Re: Probability - Siblings in the room [#permalink]

Show Tags

21 Jun 2011, 22:49

siddhans wrote:

VeritasPrepKarishma wrote:

Alchemist1320 wrote:

can we ever expect so tough questions in GMAT ?

wondering if I can ever get it

Anyways thanks for the all the replies...helps in devloping line of thinking

The question is not very difficult. Try to take one line at a time to reason it out. I will tell you what I mean. I read the first line:

"In a room filled with 7 people, 4 people have exactly 1 sibling in the room and"

I stop here. 4 people have exactly 1 sibling. So I say to myself, "Ok. A is there and he has a sibling B. Wait, this means that automatically B has the sibling A. They cannot have any other sibling since both should have only one sibling each. Then out of 4 people, 2 are already accounted for. Then in the same way, there must be C there and his sibling D". The important thing to realize is that 'sibling' relation is symmetric. If A is sibling of B, B is also the sibling of A.

Next line of the question: "3 people have exactly 2 siblings in the room." Now I think, "There is E there and he has two siblings F and G. So automatically F and G both have 2 siblings each too."

This gives me following siblings: A - B C - D E - F - G

Now I need to pick 2 people such that they are not siblings. I can do it in 2 ways. I can either find the number of ways of picking siblings or number of ways of picking 'non siblings'. Number of ways of picking siblings is certainly easier - I can pick A-B or C-D or 2 of E-F-G in 3 ways (E-F or F-G or E-G) so there are a total of 5 ways of picking siblings. Total ways of picking 2 people is 7C2 = 21 ways Out of these 21 ways, 5 ways are there to pick 2 siblings so other 16 ways must be of picking 'non siblings'.

Required probability = 16/21

Can someone please explain this ?

I can pick A-B or C-D or 2 of E-F-G in 3 ways (E-F or F-G or E-G) so there are a total of 5 ways of picking siblings. Total ways of picking 2 people is 7C2 = 21 ways Out of these 21 ways, 5 ways are there to pick 2 siblings so other 16 ways must be of picking 'non siblings'.

Required probability = 16/21

And also, How do we get 7C2?? Are there 7 possible cases?

A-B C-D E-F F-G E-G[/quote]

7C2 because there are 7 people in the room and you need to select 2 individuals..

Re: Probability - Siblings in the room [#permalink]

Show Tags

21 Jun 2011, 23:03

siddhans wrote:

agdimple333 wrote:

Other way - Probability that 2 individuals chosen are siblings- from 3 groups , A- B, C-D, E-F-G 2 * 1 / 7 * 6 + 2 * 1 / 7 * 6 + 3 * 2/ 7 * 6 = 10 / 7 * 6 = 5/21 Are not siblings = 1-5/21 = 16/21

Hi sudhir,

Can you explain this method too?

okai .. here it goes, the siblings are A-B, C-D and E-F-G selecting 2 siblings from a,b : first can be done in 2/7 ways , second in 1/6 = 2/7*1/6 = 2/42 selecting 2 sibling from c,d : first in 2/7 ways , second in 1/6 = 2/42 selecting 2 siblings from e f g : first in 3/42 ways second in 2/42 ways = 6/42 ways

total ways to select siblings = 2/42+2/42+6/42 = 10/42 = 5/21

Re: Probability - Siblings in the room [#permalink]

Show Tags

21 Jun 2011, 23:09

sudhir18n wrote:

siddhans wrote:

agdimple333 wrote:

Other way - Probability that 2 individuals chosen are siblings- from 3 groups , A- B, C-D, E-F-G 2 * 1 / 7 * 6 + 2 * 1 / 7 * 6 + 3 * 2/ 7 * 6 = 10 / 7 * 6 = 5/21 Are not siblings = 1-5/21 = 16/21

Hi sudhir,

Can you explain this method too?

okai .. here it goes, the siblings are A-B, C-D and E-F-G selecting 2 siblings from a,b : first can be done in 2/7 ways , second in 1/6 = 2/7*1/6 = 2/42 selecting 2 sibling from c,d : first in 2/7 ways , second in 1/6 = 2/42 selecting 2 siblings from e f g : first in 3/42 ways second in 2/42 ways = 6/42 ways

total ways to select siblings = 2/42+2/42+6/42 = 10/42 = 5/21

ways of not selecting sibling = 1-5/21 = 16/21

You mean selecting 2 siblings from 7 people can be done in 2/7 ways and after selecting one sibling there are 6 people left so selecting 1 from them will be 1/6 ??? I am still not clear about 2/7 and 1/6?? Sorry

Also, you take combinations when order doesnt matter ? Correct? so in 7c2 order didnt matter A-C or C-A is same?

Re: Probability - Siblings in the room [#permalink]

Show Tags

22 Jun 2011, 00:16

Dont get confused by multiple approaches mentioned in this thread. We can use both permutations and combinations. 7C2- was using combination approach. but in the above example we have used permutations.. totsal is 7P2 = 42.

You mean selecting 2 siblings from 7 people can be done in 2/7 ways and after selecting one sibling there are 6 people left so selecting 1 from them will be 1/6 ??? I am still not clear about 2/7 and 1/6??

yes ur right !! selecting one out of six but that one should be from the already slected pair of siblings. So once we select a in 2/7 ways , we can only chose b (no one else) in 1/6 ways

gmatclubot

Re: Probability - Siblings in the room
[#permalink]
22 Jun 2011, 00:16

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

They say you get better at doing something by doing it. then doing it again ... and again ... and again, and you keep doing it until one day you look...