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

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

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sm021984 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 are a total of 9 dogs - 3 have exactly 2 litter mates. 6 have only 1 litter mate.
Say A has two litter mates B and C. Then B also has two litter mates - A and C and C also has two litter mates A and B. There will be only one such group of 3 dogs where each has exactly 2 litter mates.
Of the remaining 6, say D's litter mate is E. Then E's litter mate is D. Neither of them will have any other litter mate.
Similarly, F's litter mate is G and G's litter mate is F.
H's litter mate is I and I's litter mate is H.

A, B, C
D, E
F, G
H, I

In how many ways can we select two dogs such that they ARE litter mates?
You can do it in two ways:

Finding number of combinations:
We can select D, E or F, G or H, I i.e. 3 ways.
Of A, B, C, we can select any two in 3C2 = 3 ways.
Total ways of selecting 2 litter mates is 3+3 = 6

Total ways of selecting 2 dogs out of 9 is 9C2 = 36

Probability of selecting 2 dogs such that they are litter mates = 6/36 = 1/6
Probability of selecting 2 dogs such that they are not litter mates = 1 - 1/6 = 5/6

Directly using probability:

Probability of selecting a dog out of D, E, F, G, H and I = 6/9 = 2/3
Probability of selecting his litter mate = 1/8

Probability of selecting a dog out of A, B and C = 3/9 = 1/3
Probability of selecting his litter mate = 2/8 = 1/4

Probability of selecting 2 litter mates = 2/3 *1/8 + 1/3*1/4 = 1/6
Probability of selecting non litter mates = 1 - 1/6 = 5/6
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]

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New post 14 Jul 2011, 08:29
what is the meaning of a litter mate btw and what if on the test day i get this kind of term which i dont understand like a dogs littermate . :?
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garimavyas wrote:
what is the meaning of a litter mate btw and what if on the test day i get this kind of term which i dont understand like a dogs littermate . :?


Yeah, I wasn't thrilled when I came across this word either. But it isn't hard to guess who a litter mate is. A litter mate would be a dog born in the same litter. I really doubt you will get any ambiguous terminology on the real test.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]

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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

Last edited by Bunuel on 06 May 2012, 02:29, edited 1 time in total.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]

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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 have exactl [#permalink]

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

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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 have exactl [#permalink]

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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 have exactl [#permalink]

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

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New post 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!

But this doesn't seem to work?

Bunuel please help

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

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New post 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!

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. 6 of the dogs have exactl [#permalink]

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New post 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!

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. 6 of the dogs have exactl [#permalink]

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New post 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!

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. 6 of the dogs have exactl [#permalink]

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New post 03 Nov 2012, 17:01
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 have exactl [#permalink]

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New post 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.

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

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

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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 have exactl [#permalink]

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

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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl   [#permalink] 26 Feb 2014, 22:40

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