Probability problem 2 There are 10 boxes each containing 6 white and 7 red balls. Two random
boxes are chosen, one ball is drawn simultaneously at random
from each and transferred to the other box. Now a box is again chosen
from the 10 boxes and a ball is chosen from it. Then the probability
that this ball is white is 
(A) 6/13 (B) 7/13 (C) 5/13 (D) none of these.
My attempt: There are 3 cases of transfer 1) White White Transfer
2)Red-Red Transfer
3)White Red Transfer
The probability for the cases 1 & 2 will be (1/10)(6/13). 
The probability for 3rd case is 2/10((7/13)+(5/13))+(1/8)*(6/13)
Total Probability= Probality for case 1+ Probability for case 2+ Probability for case 3 = 0.33 which comes to none of this.The correct answer is A) 6/13 .
Where I am going wrong.I think I cannot evaluate the probability for 3rd case correctly.

Hint:
There are $130$ balls in total that have equal chances to be drawn. $60$ of them are white.
