You have 6 red balls, 6 blue, and 6 white. You randomly select a sample of 3 balls without replacement. You have 6 red balls, 6 blue, and 6 white. You randomly select a sample of 3 balls without replacement. The balls, save for the color, are indistinguishable from each other. What are the odds that your sample contains at least a ball from each color?
I thought about finding the chance that the sample will not contain at least one color and subtracting that from one:
$$P(\text{sample is missing at least one color})=P(\text{sample is missing one color})+P(\text{sample is missing two colors}) $$
Essentially the second probability statement on the right hand side is the probability that all the balls in the sample are the same color. The case where the sample does not contain three colors doesn't make sense, so I didn't include it.
$$=\binom{3}{1} \left[\left(\frac{12}{18}\right) \cdot \left(\frac{11}{17}\right) \cdot\left(\frac{10}{16}\right)\right]+\binom{3}{2}\left[\left(\frac{6}{18}\right) \cdot \left(\frac{5}{17}\right) \cdot\left(\frac{4}{16}\right)\right] \approx 0.882353$$
When all of this is done, we simply subtract the probability from one to get $0.117647$.
What am I doing wrong?
 A: Method 1:  We find the probability that the three balls are each of a different color.
There are $\binom{18}{3}$ ways to select a subset of three of the eighteen balls.  From these, we select one of six blue balls, one of six white balls, and one of six white balls.  Hence, the probability that the three balls we select are each of a different color is 
$$\frac{\dbinom{6}{1}\dbinom{6}{1}\dbinom{6}{1}}{\dbinom{18}{3}}$$
Method 2:  We find the probability that at least one color is missing and subtract that from $1$.
There are $\binom{3}{1}$ ways to exclude one of the colors and $\binom{12}{3}$ ways to select three of the remaining twelve balls.
However, we have counted those selections in which all the balls are of the same color twice, once for each of the ways we could exclude one of the other two colors.  Since we only want to exclude these selections once, we must add the number of selections in which only one color is selected.  There are $\binom{3}{2}$ ways to exclude two colors and $\binom{6}{3}$ ways to select three balls of the remaining color.  
Hence, the probability that at least one color is missing is 
$$\frac{\dbinom{3}{1}\dbinom{12}{3} - \dbinom{3}{2}\dbinom{6}{3}}{\dbinom{18}{3}}$$
Subtracting this from $1$ gives the probability that a ball of each color is selected.
A: Notice in your formula 
$$P(\text{sample will not contain at least one color})$$
$$=P(\text{sample doesn't contain one color})$$$$+P(\text{sample doesn't contain two colors}) $$
The two probability you try to decompose are not mutually exclusive.
sample doesn't contain one color: contains two colors or $\color{red}{\text{three colors}}$.
sample doesn't contain two colors: contains one color or $\color{red}{\text{three colors}}$.
EDIT
I'll provide a method that follows exactly the intended/original carve-out of the OP.
$$P(\text{sample is missing contain one color})$$
$$=P(\text{sample has exactly two colors})$$
$$=\frac{\dbinom{3}{2}\times \dbinom{2}{1} \times \dbinom{6}{1} \times \dbinom{6}{2}}{\dbinom{18}{3}} $$
$$P(\text{sample is missing two colors})$$
$$=P(\text{sample has exactly one colors})$$
$$=\frac{\dbinom{3}{1}\times \dbinom{6}{3}}{\dbinom{18}{3}} $$
A: Your probability for the 2 color case is wrong. To have one color missing, you need to have one of the next two draws not match the first one (12 ways) and one of them match (5 ways), or to have them match each other but not the first (12 ways for the second and 5 ways for the third). This gives
$$
P(\mathrm{1\;color\;missing}) = 3\frac{12\cdot 5}{17\cdot 16} = \frac{45}{68}
$$
