Probability You Choose at least one chip of every color There are 16 chips: 6 red, 7 white, and 3 blue. 4 chips are selected randomly and are not replaced once selected.
What is the probability that at least one chip of every color is selected?
I'm not every good at figuring out how to create an equation of probability for these "at least one of each kind" problems. Thanks for your help and kindness in advance. It is much appreciated. 
 A: Denoting the $(N_r, N_w, N_b)$ the triple of counts of each color in the sample, configurations that meet the requirements are $(2,1,1)$, $(1,2,1)$ or $(1,1,2)$.
The probability of the first configurations is 
$$
   p_1 = \frac{\binom{6}{2} \cdot \binom{7}{1}\cdot \binom{3}{1}}{\binom{16}{4}} = \frac{9}{52}
$$
Compute $p_2$ and $p_3$ similarly. Since these events are disjoint the answer is 
$$
   p_1 + p_2 + p_3 = \frac{9}{20}
$$
The tuple $(N_r, N_w, N_b)$ follows multivariate hypergeometric distribution, and you seek evaluate 
$$
   \Pr\left(N_r > 0 \land N_w > 0 \land N_r > 0 \right)
$$
A: If you get at least one chip of every color, there are 3 cases to consider:
1) 2 red, 1 white, 1 blue
2) 1 red, 2 white, 1 blue
3) 1 red, 1 white, 2 blue.
Therefore the probability is given by 
$\displaystyle\frac{\binom{6}{2}(7)(3)+\binom{7}{2}(6)(3)+\binom{3}{2}(6)(7)}{\binom{16}{4}}=\frac{9}{20}$.
A: Since you're just learning this stuff, maybe this will help.
One thing I try to keep in mind when dealing with "at least one of" type problems: 

    Call "p" the probability of getting NO occurrences of whatever it is (basically all the ways things can be chosen so that there are no occurrences of the target item).  

    Then (1-p) is the probability of getting is the probability of getting at least one (or more).

