Drawing at least one of each color marble Suppose I have 300 marbles. 24 of them are red, 59 of them are green, 66 of them are blue, and 151 of them are yellow. What's the probability of drawing at least one of each color after 30 draws?
So I know that the $P($at least 1 red marble$) = 1 - P($no red marble$)$. And the same goes for the other colors. In that case,
$P($at least 1 red marble$) = 1 - P($no red marble$) = 1 - \frac{\binom{276}{30}}{\binom{300}{30}} = 0.928$
$P($at least 1 green marble$) = 1 - P($no green marble$) = 1 - \frac{\binom{241}{30}}{\binom{300}{30}} = 0.999$
$P($at least 1 blue marble$) \approx 1$
$P($at least 1 yellow marble$) \approx 1$
If I did the math right, I am a little surprised by the result. Even if only $24/300 = 0.08$ (i.e, only 8% of the marbles are red), the probability of drawing at least 1 red marble in a sample of 30 is still very high (0.928)? I would've thought it would be lower because red marbles are present at a very low frequency in the population. Is this because I am drawing quite a large sample (30) so the probability of drawing at least one of the rarest marble is high?
Also, to answer the question of what's the probability of drawing at least one of each color, do I have to multiply the 4 probabilities? i.e
$0.929 * 0.999 * 1 * 1 = 0.927$ 
Edit: To answer the above question I will use the hypergeometric distribution
h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ]
N: The number of items in the population. (300 in my case)
k: The number of items in the population that are classified as successes.
n: The number of items in the sample. (30 in my case)
x: The number of items in the sample that are classified as successes.
kCx: The number of combinations of k things, taken x at a time.
I'm interested in what's the probability of drawing at least one of each of the 4 color marbles given 30 draws from a population of 300 marbles. But I'm confused as to how to use the hypergeometric distribution to answer this question
 A: a :  the number of red marbles
b :  the number of green marbles
c :  the number of blue marbles
d :  the number of yellow marbles

For each vector [$a,b,c,d$] with $a+b+c+d=30$, the probability that the number
of drawn marbles of each colour is $a,b,c,d$ respectively is
$$\frac{\binom{24}{a}\binom{59}{b}\binom{66}{c}\binom{151}{d}}{\binom{300}{30}}$$
This is also true for $a>24$ because $\binom {m}{n}$ is defined to be $0$,
if $m<n$.
For the probability that EACH colour has been drawn at least once, you have
to add the probabilities for all vectors for which $min(a,b,c,d)>0$ holds.
For the probability, that lets say, at least one green marble is drawn,
you have to add all the probabilities for all vectors with $a>0$ (But
you have shown that there is an easier way to calculate this).
A: Here are two solutions, one using inclusion/exclusion and one using generating functions.  The first of these is suitable for calculation with a pocket calculator or spreadsheet, i.e. a computer program is not required, although I suppose we might regard a spreadsheet as a form of program.
The number of collections of 30 marbles taken from the set of 300 is $N=\binom{300}{30}$, each of which we assume is equally likely.  It will simplify notation a bit if we number the colors 1 through 4 and let the number of marbles of color $i$ be $n_i$, so $n_1=24, n_2=59, n_3=66$ and $n_4=151$.  Let's say a collection of marbles has "property $i$" if it has no marbles of color $i$.  We would like to count the number of collections which have none of the four properties, and we can do this with the Principle of Inclusion / Exclusion (PIE).  Let $S_i$ be the number of collections with $i$ of the properties, for $i=1,2,3$. Then
$$S_1 = \sum_{i=1}^4 \binom{300-n_i}{30}$$
$$S_2 = \sum_{i<j} \binom{300-n_i-n_j}{30}$$
$$S_3 = \sum_{i<j<k} \binom{300-n_i-n_j-n_k}{30}$$
By PIE, the number of collections with none of the properties is
$$N_0 = N - S_1 + S_2 - S_3 \approx 1.60573 \times 10^{41}$$
and the probability of having at least one marble of each color is $p = N_0 / N \approx 0.927133$.
Note: we regard a binomial coefficient of the form $\binom{n}{m}$ with $n < m$ as zero.  We encounter one of these coefficients in the computation of $S_3$.

Now for a solution using generating functions.  We ask a more general question: What is the number of collections of $r$ marbles with at least one marble of each color?  Let's call that number $a_r$ and define 
$$f(x) = \sum_{r=0}^{\infty} a_r x^r$$  Our previous solution was the case $r = 30$.  It becomes apparent after a bit of thought that
$$f(x) = \prod_{i=1}^4[(1+x)^{n_i} - 1]$$  It's possible to "read off" the coefficient of $x^{30}$ in the expansion of $f(x)$, yielding the same calculation as with PIE, but it's easier to use a computer algebra system, with the result $a_{30} \approx 1.60573 \times 10^{41}$, which agrees with our calculation of $N_0$ above.
