Consider 1-st draw of the ball from Box1 to Box2.

With the probability 4/6 you move the white ball, and with probability 2/6 you move the black box.

So, after 1-st draw 

    a) with probability 4/6 we will have (4W,2B) in Box2, and 

    b) with probability 2/6 we will have ((3W,3B) there.



Now we consider 2-nd draw of the ball from Box2 to Box3. 

    We consider case a) first. It creates two sub-cases:

      c) with probability 4/6 we move white ball and then we have then (2W,2B) in Box3;

      d) with probability 2/6 we move black ball and then we have then (1W,3B) in Box3;


    Next consider case b). It creates two sub-cases:

      e) with probability 3/6 we move white ball and then we have then (2W,2B) in Box3;

      f) with probability 3/6 we move black ball and then we have then (1W,3B) in Box3.


Now we analyse each and every of these chains of moves from B1 to B2 and then from B2 to B3 with their probabilities 
to get what we will have at the end of each chain in Box 3.


      The chain a) --> c) gives (2W,2B) in Box3 with probability  = ;   (1)

      The chain a) --> d) gives (1W,3B) in Box3 with probability  = ;    (2)

      The chain b) --> e) gives (2W,2B) in Box3 with probability  = ;    (3)

      The chain b) --> f) gives (1W,3B) in Box3 with probability  = .    (4)


Now the probability to get white ball from B3 is

      P = .

In this formula, the first factor in each addend is (1),(2),(3),(4);  the second factor comes from the content of Box3.


Finally,  P =  =  = .


It is the ANSWER to question (A) of the problem.