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.

First, let's look at the transfers.
$P(WW) = P(W)P(W) = (\frac{6}{13})^2$
$P(RR) = P(R)P(R) = (\frac{7}{13})^2$
$P(WR) = P(W)P(R) = (\frac{6}{13})(\frac{7}{13})$
$P(RW) = P(R)P(W) = (\frac{6}{13})(\frac{7}{13})$
These sum to 1, so we're in good shape. 
After this transfer step, we either have 


*

*10 boxes with $6$ white and $7$ red balls (since the WW and RR transfers don't change anything) with Probability $P(WW)+P(RR)$

*8 boxes with $6$ white and $7$ red balls, 1 box with $7$ white and $6$ red balls, 1 box with $5$ white and $8$ red balls with Probability $P(WR)+P(RW)$
These are the two scenarios you are sampling from. Your final probability of getting white $P$ is now:
$P = P($scenario 1$)P($selecting white in scenario 1$) + P($scenario 2$)P($selecting white in scenario 2$)$.
It's clear that in scenario 1, that since each box is the same the probability of selecting white is $\frac{6}{13}$. Calculating scenario 2, you should find that it will also come out to $\frac{6}{13}$!
Thus your end result will be $\frac{6}{13}$, and the transfer, in fact, made no difference (see hint from @drhab).
