Solving Probability Questions: 2 Homes, Cats, Dogs

[miscellaneous_]

Hey, I think this is a combinatorial question and I'm too far removed from stats to remember. I thought it was originally a binomial distribution but then that didn't make sense after I calculated it.

What is the probabilty of 2 homes with 2 cats and 1 dog and 2 homes with 3 dogs?

How would I solve this type of question in general. Thanks.

 

[Simon Bridge]

Welcome to PF;
The question cannot be answered without more context.
What is the known information?

 

[miscellaneous_]

Thanks Simon,
There are 4 cats and 8 dogs in total.
The probability of picking a cat is simply 1/3 and the probability of picking a dog is 2/3 initially (i.e. one animal is not inherently more probable to be chosen).
There is no replacement after picking an animal.
Order does not matter.
There is nothing else that I can think of. The question could be anything, 2 teams with 2 boys and 1 girl and 2 teams with 3 girls. Basically dividing the subset of the population into specific groups.
Is there anything specific I'm not thinking about because I feel like a moron.
I hope there is a really complex answer ... it's probably not.

The total number of possibilities is 12C4 right?

 

[Simon Bridge]

OK - so houses are twice as likely to want a dog as a cat.

Guessing:
There are two homes - they take turns picking until each has three pets and each pick is independent?

If the guess is true, then:
You cannot run out of one animal for the selections of interest, so it does not matter how many dogs and cats are in the population: the probabilities won't change.

Each home with combination cat-cat-dog (any order)
What is the probability of one home picking cat-cat-dog?