Probability task (Find probability that the chosen ball is white.) I have this task in my book:
First box contains $10$ balls, from which $8$ are white. Second box contains $20$ from which $4$ are white. From each box one ball is chosen. Then from previously chosen two balls, one is chosen. Find probability that the chosen ball is white. 
The answer is $0.5$. Again I get the different answer:
There are four possible outcomes when two balls are chosen:
$w$ -white, $a$ - for another color
$(a,a),(w,a),(a,w),(w,w)$.
Each outcome has probability:
$\frac{2}{10} \cdot \frac{16}{20};  \frac{8}{10} \cdot \frac{16}{20}; \frac{2}{10} \cdot \frac{4}{20}; \frac{8}{10} \cdot \frac{4}{20};$ 
In my opinion the probability that the one ball chosen at the end is white is equal to the sum of last three probabilities $\frac{8}{10} \cdot \frac{16}{20} + \frac{2}{10} \cdot \frac{4}{20} + \frac{8}{10} \cdot \frac{4}{20}=\frac{21}{25}$. Am I wrong or there is a mistake in the answer in the book?

The desired probability is 
$$\frac{8}{10}\frac{16}{20}\frac{1}{2}+\frac{2}{10}\frac{4}{20}\frac{1}{2}+\frac{8}{10}\frac{4}{20}$$
You will notice that in $(a, w)$ or $(w, a)$ cases, there is only $\frac12$ chance of selecting the white ball.
