Probability of taking at least one ball of specific color, no reposition, order doesn't matter 
A bag contains 9 balls. 4 are red, 3 are white and 2 are yellow. Three
  balls are extracted in a succession, without reposition.
What is the probability of extracting at least 1 yellow ball?

I did:


*

*probability of 1 yellow ball: 


(A is yellow and B is other color)
$$p(ABB) = \frac{2}{9}$$
$$p(BAB) = \frac{2}{8}$$
$$p(BBA) = \frac{2}{7}$$


*

*total = $\frac{2}{9}+\frac{2}{8}+\frac{2}{7}=\frac{191}{252}$


$$\\$$


*

*probability of getting 2 yellow balls:
$$p(AAB)=\frac{2}{9}*\frac{1}{8} = \frac{1}{36}$$
$$p(ABA)=\frac{2}{9}*\frac{1}{7} = \frac{2}{63}$$
$$p(BAA)=\frac{2}{8}*\frac{1}{7} = \frac{1}{28}$$

*total = $\frac{1}{36}+\frac{2}{63}+\frac{1}{28}=\frac{2}{21}$
So the probability of getting at least 1 yellow ball is $$\frac{2}{21}+\frac{191}{252} = \frac{215}{252} = .85%$$
But my book says the solution is about 58.3%.
What did I do wrong? How do I solve this?
 A: As a general principle, when you see a question that refers to "at least (a Certain Number)", consider instead answering the complementary question "less than (a Certain Number)".  The calculation is often easier, and needs only be subtracted from $1$, to yield the solution to the original problem.(As pointed out by @Zar)
In this case the probability of missing the two yellow balls on the first draw from the $9$ in the bag is $\frac79$.
The probability of then missing the $2$ yellows in the $8$ remaining on the second draw is $\frac68$
The probability of then missing the $2$ yellows in the $7$ remaining on the third draw is $\frac57$
The probability of these three draws missing the yellows is then $$P_{Miss}=\frac{7\times 6 \times 5}{9\times 8 \times 7}=\frac5{12}$$So $$P_{Hit}=\frac7{12}$$
A: There are $\binom{7}{3}$ ways of extracting 3 balls that are not yellow, and $\binom{9}{3}$ possibile extractions. So the probability of extracting 3 non-yellow balls is
$$
p=\frac{\binom{7}{3}}{\binom{9}{3}}.
$$
Then the probability of extracting at least one yellow ball is 
$$
\overline{p}=1-p=1-\frac{\binom{7}{3}}{\binom{9}{3}}=1-\frac{5}{12}=\frac{7}{12}\approx 58.3\%.
$$
