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In a room filled with 7 people, 4 people have exactly 1 sibl [#permalink]
09 Jul 2009, 18:39
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Question Stats:
49% (02:16) correct
51% (01:00) wrong based on 282 sessions
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]
10 Jul 2009, 02:47
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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]
10 Jul 2009, 08:26
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]
10 Jul 2009, 08:48
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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]
10 Jul 2009, 09:24
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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]
16 May 2011, 04:21
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]
16 May 2011, 04:42
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]
16 May 2011, 09:07
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]
16 May 2011, 17:25
11
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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]
21 Jun 2011, 21:38
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]
21 Jun 2011, 21:49
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]
21 Jun 2011, 22:03
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]
21 Jun 2011, 22:09
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]
21 Jun 2011, 23:16
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]
21 Jun 2011, 23:16
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