<s>[INST] 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?
 [/INST] 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 
$$
\overline{p}=1-p=1-\frac{\binom{7}{3}}{\binom{9}{3}}=1-\frac{5}{12}=\frac{7}{12}\approx 58.3\%.
$$
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