Four Photos, Three Animals

A small labelled photo dataset has 12 different photos: 6 cats, 4 dogs and 2 rabbits. Before using it, you set aside a validation set of 4 photos, picked at random without replacement so that every possible group of 4 photos is equally likely.

You would like the validation set to contain at least one photo of every animal. If an animal is missing, you have no way of checking how well it is recognised.

Because every group of 4 is equally likely, the chapter's counting rule applies:

$$P(\text{event}) = \frac{\text{number of groups in the event}}{\text{total number of possible groups}}$$

To count groups, use combinations. The number of ways to choose $r$ items from $n$ different items, when the order of choosing does not matter, is

$$\binom{n}{r} = \frac{n!}{r!\,(n-r)!}$$

where $n! = n \times (n-1) \times \dots \times 2 \times 1$, and $0! = 1$. For example, $\binom{4}{2} = \frac{24}{2 \times 2} = 6$: from four photos $a, b, c, d$ the possible pairs are $ab, ac, ad, bc, bd, cd$.

What is the probability that the validation set contains at least one cat, at least one dog and at least one rabbit?

Round your answer to 4 decimal places.