TITLE: Probability of getting at least one of each balls from an urn
QUESTION [5 upvotes]: 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}$ = $.846$

REPLY [3 votes]: Comment (won't fit in 'comment' format). More compact code for simulation in R. Results agree with previous simulated and combinatorial answers (+1)s.
urn = rep(1:4, times=4)  # a number for each color
set.seed(226);  m = 10^6
u = replicate(m, length(unique(sample(urn,8))))
mean(u == 4);  2*sd(u == 4)/sqrt(m)
## 0.846082       # aprx P(all 4 colors present among 8 balls)
## 0.0007217406   # aprx 95% margin of simulation error
table(u)/m        # aprx dist'n of nr distinct colors drawn
 u
        2        3        4 
 0.000479 0.153439 0.846082  

With $m = 10^7$ iterations: $ 0.84661 \pm 0.00023\,$ on one run and
$0.84659 \pm  0.00023$ on another. Difficult to say for sure from these simulations,
but it seems the 4th decimal place is closer to 6 than to 2. Wish I had
a better intuitive understanding of the argument against $11/13$.