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$$\frac{12\choose8}{16\choose8}$ = $.846$
 A: We can use inclusion-exclusion principle:
Let $A_1,A_2,...,A_4$ denote events that there will be a red, blue, green, yellow ball in our selection respectively.
Then, by symmetry:$$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...$$
A: Python code to run a Monte Carlo simulation modeling this problem:

import numpy as np
#
urn = ['r']*4 + ['b']*4 + ['y']*4 + ['g']*4
#
numtrials = 1000000
missing_color = 0
for i in range(numtrials):
    p = np.random.choice(urn, size=8, replace=False)
    if 'r' not in p:
        missing_color += 1
    elif 'b' not in p: 
        missing_color += 1
    elif 'y' not in p: 
        missing_color += 1
    elif 'g' not in p:
        missing_color += 1
#
prob_missing = missing_color / float(numtrials)
print 'Probability of a color missing = ', prob_missing
print 'P(at least one of each ball)  = ', 1 - prob_missing

A: John H, your answer is correct, although, strictly speaking, we are dealing with outcomes here and need to consider permutations but it's gonna return the same number if we swap permutations with combinations:
$$1-4\cdot\frac{P(12,8)}{P(16,8)}\approx0.85$$
There's no guessing here. The overall number of outcomes is $P(16,8)$. 
The overall number of possible unfavorable outcomes is $\;4\cdot P(12,8)$. No guessing, your answer is correct.
You can use cards for illustration. Take 16 cards, 4 clubs, 4 spades, 4 hearts, and 4 diamonds. 
A program is, of course, a nicer option.
A: 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 $11/13$.
A: 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
