well c..

From first bag,, the white ball can be selected in  a/(a+b)  ways .. And black in b/(a+b)... Let the white be selected ,, then placed in second bag.. so second bag contains (a+b+1) for the moment,, If white is selected then it has (a+1) white balls,,, and if black is selected than it has,, (b+1) black balls[and a white] from total...

so probability of gettin white from this second bag is given by

thus selection of white from every consecutive bag (in filling process) is given by  a/(a+b)..

For the kth bag,, the selected bag can go through four cases::

1) that it has taken white from previous bag, and given white to next bag..;

2)that it has taken white from previous bag, and given black to next bag..;

3)that it has taken black from previous bag, and given white to next bag..;  and

4)that it has taken black from previous bag, and given black to next bag............

For first case,, the bag would have taken a white ball from previous bag with prob. a/(a+b)  and given white to following bag with a/(a+b) prob. the cases occur one after other..Now while selecting white ball,, the total no. of balls is (a+b) and no. of white balls is "a"...

so prob. of selection becomes::: 

since three of them occur simultaneously..

Following similar approach,, we get it for

2)

3)

4)

Adding all these possibilities we get..

bas yaar,, i could think till here!!!