<s>[INST] 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?
 [/INST] First box: $p(white) = \frac{8}{10}$
Second box: $p(white) = \frac{2}{10}$  
One of the two is chosen. If you pick the first ball $(p=0.5)$, then $p(white)=0.8$. If you pick the second $(p=0.5)$, then $p(white) = 0.2$. Now notice that p(ball 1 is white) = p(ball 2 is nonwhite) and p(ball 2 is white) = p(ball 1 is white). This symmetry lets us conclude immediately that the probability of ending up with a white ball is $0.5$.
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