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

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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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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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

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.

Merging similar topics. Please refer to the solutions above.
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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.

Regards

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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