Finding the probability of drawing all balls of the same color when there is a "wildcard" ball I am working on a practice exam question for my Probability Theory class and am confused about a solution my professor gives us.
The problem is as follows:
There is a box with 16 balls in it, 4 red, 4 green, 4 blue, and 4 rainbow. If we pick at random 5 balls without replacement, what is the probability that all 5 balls are of the same color, if we allow the rainbow balls to take any color?
I know that the sample space would be $16 \choose 5$. I also know that the probability of this would generally be $P(\text{4 red balls, 1 rainbow}) + P(\text{4 green balls, 1 rainbow}) + P(\text{4 blue balls, 1 rainbow})$. However, I'm not quite sure how to actually count all of these.
My initial thinking is to use the conditional approach and do something along the lines of:
$\frac{4}{16} * \frac{3}{15} * \frac{2}{14} * \frac{1}{13} * \frac{4}{12}$
Essentially, the probability of picking 1 ball of a specific color multiplied by the probabilities of picking another ball of that same color, and then finally multiplied by the probability of picking 1 of the rainbow balls. This doesn't seem right to me, however as I feel like I need to account for the fact that you could pick a rainbow ball at any point in there.
Could someone guide me on this? Thank you!

if you want to find the probability P(4 balls of same color and 1 rainbow colored ball) then it would be
$$ \frac{4}{16} \times \frac{3}{15} \times \frac{2}{14} \times \frac{1}{13} \times \frac{4}{12} \times \frac{5!}{4} \times 3  = 0.002747...$$
we multiply by 5!/4 if we don't care about the order in which we pick those balls(there would be 5 different permutations, rainbow first then colored ones or rainbow second and so on) and we multiply by 3 because there are three colors (excluding rainbow) you could also just add it three times.
If the rainbow colored balls can be considered of any color then the answer could be different for example if we pick 3 red and 2 rainbow colored balls does that also count as all red? if not then I believe the above answer is correct.
Edit: if for example 2 rainbow balls and 3 red balls also count as all red then the answer provided by @knrumsey is correct one, if sets like this don't count then the answer provided above is correct
