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?

Denote that event that eventually a white ball is drawn by $W$.
Denote that event that a white ball is drawn from the $i$-th box
by $W_{i}$ ($i=1,2$).
Denote that event that not a white ball is drawn from the $i$-th
box by $A_{i}$ ($i=1,2$).
Then $P\left(W\right)=P\left(W|W_{1}\cap W_{2}\right)P\left(W_{1}\cap W_{2}\right)+P\left(W|W_{1}\cap A_{2}\right)P\left(W_{1}\cap A_{2}\right)+P\left(W|A_{1}\cap W_{2}\right)P\left(A_{1}\cap W_{2}\right)+P\left(W|A_{1}\cap A_{2}\right)P\left(A_{1}\cap A_{2}\right)$
Here $P\left(W_{1}\cap W_{2}\right)=P\left(W_{1}\right)\times P\left(W_{2}\right)$,
$P\left(W_{1}\cap A_{2}\right)=P\left(W_{1}\right)\times P\left(A_{2}\right)$..
et cetera because of independency. 
This leads to: $P\left(W\right)=1\times\frac{8}{10}\times\frac{4}{20}+\frac{1}{2}\times\frac{8}{10}\times\frac{16}{20}+\frac{1}{2}\times\frac{2}{10}\times\frac{4}{20}+0\times\frac{2}{10}\times\frac{16}{20}=\frac{1}{2}$.
