ok..here's a nice Q........there are n urns.....each urn contains 'a' white and 'b' black balls........i first pick up one ball at random from the first urn and place it inside the second urn.....then pick up a ball at random frm the second urn and put it into the 3rd urn........and pick up a ball at random frm the 3nd urn and put it into the 4th urn and so on.......after the whole process is over....wats the probability that a randomly picked up ball from the 'k'th urn turns out to be a white ball.........

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!!!

 jus a small silly mistake.....but i guess u shud continue with da soln..............

is it wrong?/

if then pls. correct it..

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 '..........

ooopsss........

sorry yaar..my soln. is totally wrong...........crappp donot refer it.............forgot the main part of the problem........

experts.....frnds help us out of this one..........