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

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