Probability Urn Question(Recurrence Model) From an urn containing a white and b black balls, balls are drawn one
by one at random according to the following rules:

(i) at any drawing, if the ball drawn is white, then it is returned to the urn,

(ii) if it is black, it is replaced by a white ball (from another collection of
balls).

After n such operations, a ball is drawn from the urn. Find the probability
that it will be white.

my Approach:
Let w(k) denote that ball from kth selection is white,k = 1,2,...,n

suppose p_k = P(w(k)),we have to find p_n = P(w(n))..
i know surely that,we have to obtain Recurrence relation,but don't know how to manipulate p_n.

Correct Answer:$1-($($b/(a+b)$)$(1-1/(a+b))^{n}$)
There are $w+b$ balls.  Number them from $1$ to $w+b$.  When a black ball is replaced, give the new white ball the same number.
For the ball to be black on the $k^{th}$ draw, it must be the first time that number was drawn.  The number of ways to do that is $(w+b-1)^{k-1}$.
Can you finish the question?
At each draw when a black ball is drawn the probability of selection of white increases in the consecutive  draw.
We can have the following $n+1$ different combinations of drawings of $n$-balls,(where $w_k$ represents a white ball is drawn on $k$th draw and $B_k$ represents a black ball is drawn at $k$th draw) after which we have  to obtain the probability of getting a white ball.


$w_1w_2w_3w_4\dots  w_n$                 
$w_1w_2w_3w_4 \dots B_n$
$w_1w_2w_3w_4\dots B_{n-1}B_n$


and so on...

n)     $w_1B_2B_3B_4\dots B_{n-1}B_n$

n+1)   $B_1B_2B_3B_4\dots B_{n-1}B_n$

The probability of getting the white ball in $n+1$th drawing by 1) will be $a/(a+b)$
by 2) will $(a+1)/(a+b)$ and so on.............. by n+1) will be $(a+n)/(a+b)$

Thus the required probability should be $(a+a+1+a+2+........+a+n)/(a+b)$ (since each of the ways are independent of each other.