Probability of getting at least one of each balls from an urn Consider an urn containing $4$ red balls, $4$ blue balls, $4$ yellow balls, and $4$ green balls. If 8 balls are randomly drawn from these 16 balls, what is the probability that it will contain at least one ball of each of the four colors?

Attempted Solution:

I think I did this right, but I just wanted to confirm.

P(at least one of each ball) 

= $1$ - P(not one of each)

= $1$ - $\frac{12\choose8}{16\choose8}$ = $.9615$

Edit:

Actually, I think I have to choose one group of 4 to not get any selected, i.e. $4\choose1$ and then take 

$1$ - $4$$\frac{12\choose8}{16\choose8}$ = $.846$
We can use inclusion-exclusion principle:
Let $A_1,A_2,...,A_4$ denote events that there will be a red, blue, green, yellow ball in our selection respectively.

Then, by symmetry:$$P(A_1^c\cup...\cup A_4^c) = 4P(A_1^c)-6P(A_1^c\cap A_2^c) + 4P(A_1^c\cap A_2^c \cap A_3^c) - P(A_1^c\cap A_2^c \cap A_3^c \cap A_4^c) $$

Noting that $$P(A_1^c) = \frac{\binom{12}{8}}{\binom{16}{8}} \quad \quad P(A_1^c\cap A_2^c) = \frac{\binom{8}{8}}{\binom{16}{8}}$$ and the remaining terms are zeroes.

Thus $$P(A_1 \cap...\cap A_4) = 1-P(A_1^c\cup...\cup A_4^c) = \color{red}{\frac{1816}{2145}}=0.846620...$$

which is different from your proposed answer:
$$1-4\frac{\binom{12}{8}}{\binom{16}{8}} = \frac{11}{13} = 0.846154...$$
OP answer and method is not correct. 

still does not count correctly.

this can be easily seen if draws 4 thru 13 are done (the distribution)

(left to the reader of course)

-Inclusion-exclusion method gets it right-

I used this:

(C(16,8)^-1) * sum((-1)^k * C(4,k) * C(4*(4-k),8)) , k=0 to 3

in Wolfram Alpha

returns
1816/2145

or
≈0.84662

pisco125 answer exactly

Sally

added:
using brute force counting in Excel and using some R code (not shown)

I get 85 unique draw results (combinations)
where 31 have at least 1 (complete set) of each color

adding the number of ways for each unique combination...

that fraction is
10896/12870

Excel snapshot

that reduces to 1816/2145
as before