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A dog breeder currently has 9 breeding dogs. 6 of the dogs

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A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 05 May 2012, 18:55
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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. 1/6
B. 2/9
C. 5/6
D. 7/9
E. 8/9
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 May 2012, 01:29, edited 1 time in total.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 06 May 2012, 01:14
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BhaskarPaul 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


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;

P=1-6/36=5/6.

Answer: C.

Similar question to practice: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html

Hope it helps.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 08 May 2012, 09:50
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1) select no littermates from Group 1 (1-2), (3-4), (5-6)
6*2=12 (i.e. 1-3; 1-4; 1-5;1-6;2-3;2-4;2-5;2-6;3-5;3-6;4-5;4-6)

2) select no littermates from Group 2 (7-8-9) . it equals to zero

3) select a dog from Group 1 and one from Group 2 (again no littermate)

3*6=18

(12+0+18)/9C2=30/36=5/6
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A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 28 Jun 2012, 23:59
Well i find it easy to solve these relationship problems plainly by looking at the number of relationships.

Since 6 have exactly 1 littermate ,
what u really have are 3 relationships amongst 6 .
1->2
3->4
5->6

Since 3 have exactly 2 relationships.
what u really have are 3 relationships.
7->8
8->9
7->9

Total of 6 relationships.
Probability they will be relatied = number of relationships/ total number of ways u can pick 2 from 9
= 6/ 9C2
=6/36=1/6

Hence Probablity that they are not related is = 1- 1/6 = 5/6
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 29 Jun 2012, 17:52
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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


There is a very similar question here: probability-siblings-in-the-room-80722.html#p922063
You can use the exact same logic to arrive at the answer.

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
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 30 Jun 2012, 03:20
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
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A dog breeder currently has 9 breeding dogs. [#permalink] New post 03 Nov 2012, 06:38
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!

But this doesn't seem to work?

Bunuel please help

OA
[Reveal] Spoiler:
C
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Re: A dog breeder currently has 9 breeding dogs. [#permalink] New post 03 Nov 2012, 06:47
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!

But this doesn't seem to work?

Bunuel please help

OA
[Reveal] Spoiler:
C


How to understand 6 dogs have 1 littermate and 3 dogs have 2 littermates and the question is about dogs without littermates?
:shock:
What does littermate in this case mean?
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Re: A dog breeder currently has 9 breeding dogs. [#permalink] New post 03 Nov 2012, 06:57
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!

But this doesn't seem to work?

Bunuel please help

OA
[Reveal] Spoiler:
C


How to understand 6 dogs have 1 littermate and 3 dogs have 2 littermates and the question is about dogs without littermates?
:shock:
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.
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Re: A dog breeder currently has 9 breeding dogs. [#permalink] New post 03 Nov 2012, 07:15
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!

But this doesn't seem to work?

Bunuel please help

OA
[Reveal] Spoiler:
C


How to understand 6 dogs have 1 littermate and 3 dogs have 2 littermates and the question is about dogs without littermates?
:shock:
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? :?
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Re: A dog breeder currently has 9 breeding dogs. [#permalink] New post 03 Nov 2012, 08:38
Expert's post
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!

But this doesn't seem to work?

Bunuel please help

OA
[Reveal] Spoiler:
C


Merging similar topics. Please refer to the solutions above.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 03 Nov 2012, 16:01
HI Bunuel

Here is how i approached the problem
probablity of getting exactly two littermate = 1/6 * 2/3 = 1/9

Prob of not getting two littermate = 1 - 1/9 = 8/9

Another problem i was facing her to decide between and , or which operation to follow.

Can u pls help me with that.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 03 Nov 2012, 16:10
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.

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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 29 Nov 2012, 13:16
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, 01:22, edited 1 time in total.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 06 Dec 2012, 22:53
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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
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 08 Dec 2012, 01:34
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
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs [#permalink] New post 26 Feb 2014, 21:40
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs   [#permalink] 26 Feb 2014, 21:40
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