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}$ = $.846$

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
