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A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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05 May 2012, 19:55

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A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

A. 1/6 B. 2/9 C. 5/6 D. 7/9 E. 8/9

Since there are 6 dogs with exactly 1 littermate each, then we have three pairs of littermates: (1-2), (3-4), (5-6); Since there are 3 dogs with exactly 2 littermates each, then we have one triple of littermates: (7-8-9).

Let's find the probability of the opposite event, so the probability that both selected dogs ARE littermates, and subtract it from 1.

We can select either (1-2), (3-4), or (5-6) so 3 ways; We can select any two from (7-8-9) so \(C^2_3=3\) ways; So, total ways to select 2 littermates out of these 9 dogs is 3+3=6;

Total # of ways to select 2 dogs out of 9 is \(C^2_9=36\);

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

Try to take one line at a time to reason it out. Everything will easily fall in place. I will tell you what I mean. I read the first line:

"In a room filled with 9 breeding dogs, 6 have exactly 1 littermate and"

I stop here. 6 dogs have exactly 1 littermate. So I say to myself, "Ok. A is there and it has a littermate B. Wait, this means that automatically B has the littermate A. They cannot have any other littermate since both should have only one littermate each. Then out of 6 dogs , 2 are already accounted for. Then in the same way, there must be C and its littermate D and there must also be E and its littermate F". The important thing to realize is that 'littermate' relation is symmetric. If A is littermate of B, B is also the littermate of A.

Next line of the question: "3 of the dogs have exactly 2 littermates" Now I think, "There is G and it has two littermates H and J. So automatically H and J both have 2 littermates each too."

This gives me following littermates: A - B C - D E - F G - H - I

Now I need to pick 2 dogs such that they are not littermates. I can do it in 2 ways. I can either find the number of ways of picking littermates or number of ways of picking 'non littermates'. Number of ways of picking littermates is certainly easier - I can pick A-B or C-D or E-F or 2 of G-H-I in 3 ways (G-H or H-I or G-I) so there are a total of 6 ways of picking littermates. Total ways of picking 2 dogs is 9C2 = 36 ways Out of these 36 ways, 6 ways are there to pick 2 littermates so other 30 ways must be of picking 'non littermates'.

Required probability = 30/36 = 5/6
_________________

Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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30 Jun 2012, 04:20

Joy111 wrote:

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

A. 1/6 B. 2/9 C. 5/6 D. 7/9 E. 8/9

We have three pairs of dogs for the 6 with exactly one littermate, and one triplet, with each having exactly two littermates. So, in fact there are two types of dogs: those with one littermate - say A, and the others with two littermates - B.

Work with probabilities: Choosing two dogs, we can have either one dog of type B or none (we cannot have two dogs both of type B). The probability of choosing one dog of type B and one of type A is 3/9 * 6/8 * 2 = 1/2 (the factor of 2 for the two possibilities BA and AB). The probability of choosing two dogs of type A which are not littermates is 6/9 * 4/8 = 1/3 (choose one A, then another A which isn't the previous one's littermate). The required probability is 1/2 + 1/3 = 5/6.

Find the probability for the complementary event: choose AA or BB. Probability of choosing two dogs of type A who are littermates is 6/9 * 1/8 = 1/12. Probability of choosing two dogs of type B (who necessarily are littermates) is 3/9 * 2/8 = 1/12. Again, we obtain 1 - (1/12 + 1/12) = 5/6.

Answer: C
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

A dog breeder currently has 9 breeding dogs. [#permalink]

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03 Nov 2012, 07:38

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates ?

A)1/6 B)2/9 c)5/6 D)7/9 E)8/9

If I name the dogs A thro I then then below are the pairs ->

AB BA CD DC EF FE G H and I H G and I I G and H

To select a pair that is not a little mate then

I can select any of 9 dogs for the 1st slot....

then how to do i got about? I could i have selected A in my 1st pick then i have remaining 7 to chose from (leaving B) = 9 * 7 / 2!

but If i select G 1st the I have 6 dogs left to chose from (leaving I and H) = 9 * 6 /2!

Re: A dog breeder currently has 9 breeding dogs. [#permalink]

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03 Nov 2012, 07:47

Jp27 wrote:

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates ?

A)1/6 B)2/9 c)5/6 D)7/9 E)8/9

If I name the dogs A thro I then then below are the pairs ->

AB BA CD DC EF FE G H and I H G and I I G and H

To select a pair that is not a little mate then

I can select any of 9 dogs for the 1st slot....

then how to do i got about? I could i have selected A in my 1st pick then i have remaining 7 to chose from (leaving B) = 9 * 7 / 2!

but If i select G 1st the I have 6 dogs left to chose from (leaving I and H) = 9 * 6 /2!

Re: A dog breeder currently has 9 breeding dogs. [#permalink]

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03 Nov 2012, 07:57

actleader wrote:

Jp27 wrote:

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates ?

A)1/6 B)2/9 c)5/6 D)7/9 E)8/9

If I name the dogs A thro I then then below are the pairs ->

AB BA CD DC EF FE G H and I H G and I I G and H

To select a pair that is not a little mate then

I can select any of 9 dogs for the 1st slot....

then how to do i got about? I could i have selected A in my 1st pick then i have remaining 7 to chose from (leaving B) = 9 * 7 / 2!

but If i select G 1st the I have 6 dogs left to chose from (leaving I and H) = 9 * 6 /2!

How to understand 6 dogs have 1 littermate and 3 dogs have 2 littermates and the question is about dogs without littermates?

What does littermate in this case mean?

I just translated "littermate" to pair... Although this is from kaplan I have seen more similar questions from Manhattan, which are quite strait forward for example:

there 8 ppl in a committee each have 1 sibling pair how many ways to select a 3 ppl who are NOT a sibling pair ->

8 * 6 * 4 / 3! = 32. this 1 is extension of this problem with more pairs hence more difficult.

Re: A dog breeder currently has 9 breeding dogs. [#permalink]

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03 Nov 2012, 08:15

Jp27 wrote:

actleader wrote:

Jp27 wrote:

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates ?

A)1/6 B)2/9 c)5/6 D)7/9 E)8/9

If I name the dogs A thro I then then below are the pairs ->

AB BA CD DC EF FE G H and I H G and I I G and H

To select a pair that is not a little mate then

I can select any of 9 dogs for the 1st slot....

then how to do i got about? I could i have selected A in my 1st pick then i have remaining 7 to chose from (leaving B) = 9 * 7 / 2!

but If i select G 1st the I have 6 dogs left to chose from (leaving I and H) = 9 * 6 /2!

How to understand 6 dogs have 1 littermate and 3 dogs have 2 littermates and the question is about dogs without littermates?

What does littermate in this case mean?

I just translated "littermate" to pair... Although this is from kaplan I have seen more similar questions from Manhattan, which are quite strait forward for example:

there 8 ppl in a committee each have 1 sibling pair how many ways to select a 3 ppl who are NOT a sibling pair ->

8 * 6 * 4 / 3! = 32. this 1 is extension of this problem with more pairs hence more difficult.

it is harder because of added probability conditions. in the case with M's committees it's simply a combinatorial.

so if i've understand you rightly - thare are aditionall 6+3 littermates along with 9 dogs or I'm blunting?

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates ?

A)1/6 B)2/9 c)5/6 D)7/9 E)8/9

If I name the dogs A thro I then then below are the pairs ->

AB BA CD DC EF FE G H and I H G and I I G and H

To select a pair that is not a little mate then

I can select any of 9 dogs for the 1st slot....

then how to do i got about? I could i have selected A in my 1st pick then i have remaining 7 to chose from (leaving B) = 9 * 7 / 2!

but If i select G 1st the I have 6 dogs left to chose from (leaving I and H) = 9 * 6 /2!

Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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03 Nov 2012, 17:10

EvaJager wrote:

Joy111 wrote:

A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

A. 1/6 B. 2/9 C. 5/6 D. 7/9 E. 8/9

We have three pairs of dogs for the 6 with exactly one littermate, and one triplet, with each having exactly two littermates. So, in fact there are two types of dogs: those with one littermate - say A, and the others with two littermates - B.

Work with probabilities: Choosing two dogs, we can have either one dog of type B or none (we cannot have two dogs both of type B). The probability of choosing one dog of type B and one of type A is 3/9 * 6/8 * 2 = 1/2 (the factor of 2 for the two possibilities BA and AB). The probability of choosing two dogs of type A which are not littermates is 6/9 * 4/8 = 1/3 (choose one A, then another A which isn't the previous one's littermate). The required probability is 1/2 + 1/3 = 5/6.

Find the probability for the complementary event: choose AA or BB. Probability of choosing two dogs of type A who are littermates is 6/9 * 1/8 = 1/12. Probability of choosing two dogs of type B (who necessarily are littermates) is 3/9 * 2/8 = 1/12. Again, we obtain 1 - (1/12 + 1/12) = 5/6.

Answer: C

My doubt is when you know that first category has just 1 litter mate than how can we consider while calculating 1/8 as it has only one littermate . I suppose when it is stated as 1 littermate that doesnt mean 2 in number. if that is the case than in 2nd calculation 4 and 3 should be used not 3 and 2.

Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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29 Nov 2012, 14:16

VeritasPrepKarishma wrote:

Bunuel wrote:

kapilhede17 wrote:

[*]A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

A. 1/6 B. 2/9 C. 5/6 D. 7/9 E. 8/9

Please help me with a fundamental understanding:

I solved this Q using the reverse technique i.e.: Probability of not getting any littermates = prob of not getting (a. any of littermate doubles + b. any two of littermate triplets) = (2/9)*(7/8) + (2/9)*(7/8) + (2/9)*(7/8) + (3/9)*(6/8) = 7/12 + 1/4 = 5/6

So, the below method should work as well: = (2C1*7C1)*3 / 9C2 + (3C1*6C1) / 9C2

But, it does not, I guess there is some problem with 9C2, because if I replace the same with 9*8, the equation becomes the same as above

I am not able to point out the fundamental flaw in 9C2 method as we use it quite often in similar Qs

For eg: A 10-member student leadership committee consists of 4 juniors and 6 seniors. Exactly 6 students will be selected from this group to attend a national convention. What is the probability that at least 3 seniors are selected for the committee?

Sol: = 1- (4C4*6C2) / 10C6 = 13/14

Kindly help me to clear this fundamental flaw in my understanding. I have been banging my head over this for few hours:)

Last edited by hiteshwd on 08 Dec 2012, 02:22, edited 1 time in total.

I solved this Q using the reverse technique i.e.: Probability of not getting any littermates = prob of not getting (a. any of littermate doubles + b. any two of littermate triplets) = (2/9)*(7*8) + (2/9)*(7*8) + (2/9)*(7*8) + (3/9)*(6*8) = 7/12 + 1/4 = 5/6

So, the below method should work as well: = (2C1*7C1) / 9C2 + (3C1*6C1) / 9C2

But, it does not, I guess there is some problem with 9C2, because if I replace the same with 9*8, the equation becomes the same as above

I am not able to point out the fundamental flaw in 9C2 method as we use it quite often in similar Qs

For eg: A 10-member student leadership committee consists of 4 juniors and 6 seniors. Exactly 6 students will be selected from this group to attend a national convention. What is the probability that at least 3 seniors are selected for the committee?

Sol: = 1- (4C4*6C2) / 10C6 = 13/14

Kindly help me to clear this fundamental flaw in my understanding. I have been banging my head over this for few hours:)

Responding to a pm:

Finding littermates is much easier than finding non-littermates. There is much less confusion. Still, you can obviously work it out the other way around too.

Non littermates can be found in two ways:

1. You select one of the 3 littermates and one of the three pairs of 2 littermates. 3C1 * 6C1 = 18 ways

2. You select two of the three pairs of 2 littermates such that they belong to different litters. You pick any one of the 6 dogs in 6 ways (say you picked B) and then you have 4 options (since you cannot pick A now). Say, you picked C. There are 6*4 = 24 ways. But wait, here, you have arranged the picked dogs. You have picked BC. In a different case, you would have picked C and then B. But both these are the same. So you need to divide 24 by 2 i.e. 12 ways.

Total number of ways of picking non-littermates = 18+12 = 30

Number of ways of picking 2 dogs = 9C2 = 36

Required Probability = 30/36 = 5/6
_________________

Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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08 Dec 2012, 02:34

VeritasPrepKarishma wrote:

Responding to a pm:

Finding littermates is much easier than finding non-littermates. There is much less confusion. Still, you can obviously work it out the other way around too.

Non littermates can be found in two ways:

1. You select one of the 3 littermates and one of the three pairs of 2 littermates. 3C1 * 6C1 = 18 ways

2. You select two of the three pairs of 2 littermates such that they belong to different litters. You pick any one of the 6 dogs in 6 ways (say you picked B) and then you have 4 options (since you cannot pick A now). Say, you picked C. There are 6*4 = 24 ways. But wait, here, you have arranged the picked dogs. You have picked BC. In a different case, you would have picked C and then B. But both these are the same. So you need to divide 24 by 2 i.e. 12 ways.

Total number of ways of picking non-littermates = 18+12 = 30

Number of ways of picking 2 dogs = 9C2 = 36

Required Probability = 30/36 = 5/6

Thanks Karishma

I realized the mistake of double counting in my approach (also, edited my post, used some wrong signs. apologies for confusion)

Also, any advice or approach recommendation so that I keep my double count in check

Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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26 Feb 2014, 22:40

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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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13 May 2016, 09:09

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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink]

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17 Oct 2016, 08:50

VeritasPrepKarishma wrote:

hiteshwd wrote:

Please help me with a fundamental understanding:

I solved this Q using the reverse technique i.e.: Probability of not getting any littermates = prob of not getting (a. any of littermate doubles + b. any two of littermate triplets) = (2/9)*(7*8) + (2/9)*(7*8) + (2/9)*(7*8) + (3/9)*(6*8) = 7/12 + 1/4 = 5/6

So, the below method should work as well: = (2C1*7C1) / 9C2 + (3C1*6C1) / 9C2

But, it does not, I guess there is some problem with 9C2, because if I replace the same with 9*8, the equation becomes the same as above

I am not able to point out the fundamental flaw in 9C2 method as we use it quite often in similar Qs

For eg: A 10-member student leadership committee consists of 4 juniors and 6 seniors. Exactly 6 students will be selected from this group to attend a national convention. What is the probability that at least 3 seniors are selected for the committee?

Sol: = 1- (4C4*6C2) / 10C6 = 13/14

Kindly help me to clear this fundamental flaw in my understanding. I have been banging my head over this for few hours:)

Responding to a pm:

Finding littermates is much easier than finding non-littermates. There is much less confusion. Still, you can obviously work it out the other way around too.

Non littermates can be found in two ways:

1. You select one of the 3 littermates and one of the three pairs of 2 littermates. 3C1 * 6C1 = 18 ways

2. You select two of the three pairs of 2 littermates such that they belong to different litters. You pick any one of the 6 dogs in 6 ways (say you picked B) and then you have 4 options (since you cannot pick A now). Say, you picked C. There are 6*4 = 24 ways. But wait, here, you have arranged the picked dogs. You have picked BC. In a different case, you would have picked C and then B. But both these are the same. So you need to divide 24 by 2 i.e. 12 ways.

Total number of ways of picking non-littermates = 18+12 = 30

Number of ways of picking 2 dogs = 9C2 = 36

Required Probability = 30/36 = 5/6

By working directly with probabilities, I found non-littermates more directly and faster. Since there are 6 dogs with exactly 1 littermate each, then we have three pairs of littermates: (1-2), (3-4), (5-6): group A Since there are 3 dogs with exactly 2 littermates each, then we have one triple of littermates: (7-8-9): group B. The probability of not littermates is: 1st selection a dog from group B over 9 dogs , 2nd selection a dog from group A over the 8 remaining dogs : 3/9 x 6/8 To this, we add the probability of 1st selection a dog from group A over 9 dogs, 2nd selection any of the other dods except for its mate over the 8 remaining dogs: 6/9 x 7/8 3/9 x 6/8 + 6/9 x 7/8 = 5/6 Choice C _________________

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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs
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17 Oct 2016, 08:50

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