> 
> 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?