sorry yaar....i guess a/(a+b) is actually the rite answer............

look......

let W_i denote the event of getting a white ball from a random draw frm the ith bag an B for black ball..........

P(W₂) = P(W₁) P(W₂/W₁) + P(B₁) P(W₂/B₁)

= a/(a+b)* (a+1)/(a+b+1) + b/(a+b) *a/(a+b+1)

=a/(a+b)......

same way.......

P(W₃) = P(W₂) P(W₃/W₂) + P(B₂) P(W₃/B₂)

= a/(a+b) * (a+1)/(a+b+1) + b/(a+b)* a/(a+b+1)

again = a/(a+b).....

i guess this will keep continuing for all the values of i and is independent of ' i '..........