Probability of extracting a ball after two balls were swapped We have $2$ boxes, the first one contains $10$ white balls and $11$ black balls. The second box contain $12$ white balls and $13$ black balls.
We swap two balls between the boxes then we extract a ball from the first box. What is the probability that the ball is white?
A swap consists in taking a ball from the first box and put it in the second one, then taking a ball from the second box and put it in the first one.
$$P(\text{white ball from box1})=\frac{10}{21}$$
$$P(\text{black ball from  box1})=\frac{11}{21}$$
$$P(\text{white ball from box2})= \frac{12+1}{26}\frac{10}{21}+\frac{12}{26}\frac{11}{21}=\frac{13\cdot10+12\cdot11}{26\cdot 21}$$
$$P(\text{black ball from box2})= \frac{13}{26}\frac{10}{21}+\frac{13+1}{26}\frac{11}{21}=\frac{13\cdot 10+14\cdot 11}{26\cdot 21}$$
$$P(\text{white ball from box1 after swap})=\frac{10+1}{22}\cdot\frac{13\cdot10+12\cdot11}{26\cdot 21}+\frac{10}{22}\cdot\frac{13\cdot 10+14\cdot 11}{26\cdot 21}$$
$$=\frac{2882}{12012}+\frac{2840}{12012}=\frac{5722}{12012}=0.4763$$

Have I done it correctly? I don't know how to verify myself with this kind of problems. 
  Also, is there a trick to find the probability if there were two swaps?


The probability of ending up with an additional white ball equals:
$$\frac{11}{21} \frac{12}{26}$$
The probability of ending up with an additional black ball equals:
$$\frac{10}{21} \frac{13}{26}$$
We thus find an overall probability of:
$$\frac{11}{21} \frac{12}{26} \frac{11}{21} + \frac{10}{21} \frac{13}{26} \frac{9}{21} + \left(1 - \frac{11 \cdot 12 + 10 \cdot 13}{21 \cdot 26}\right) \frac{10}{21} = \frac{1452 + 1170 + 2840}{11466} = \frac{5462}{11466} \approx 0.4764$$
