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:$1-($($b/(a+b)$)$(1-1/(a+b))^{n}$)
 A: 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?
A: The key, as Michael said, is to notice that each black ball can be selected at most once. 
Number the black balls from $1$ to $b$ to distinguish them. If on a draw I draw black ball number $l$, it must be that all previous draws are some other ball. 
Now, a step by step guide:


*

*Given the above: what is the probability that black ball number $l$ is picked on the $k$th draw? 

*Given that there are $b$ black balls, what is the probability that a black ball is picked on the $k$th draw?

*Since picking a white ball means not picking a black ball, what is the probability of picking a white ball on the $k$th draw? 

*What is the correct $k$ to use if you have already repeated the procedure $n$ times and are about to pick another ball? 

A: 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.
A: 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 kth draw and $B_{k}$ represents a black ball is drawn at kth draw) after which we have to obtain the probability of getting a white ball.
1) $w_{1} w_{2} w_{3} w_{4}.......w_{n}$            (total permuted ways: n choose 0)
2) $w_{1}w_{2}w_{3}w_{4}.......B_{n}$            (total permuted ways: n choose 1)
3) $w_{1}w_{2}w_3w_4.......B_{n-1}B_n$        (total permuted ways: n choose 2)
so on..... ..................
n) $w_1B_2B_3B_4.......B_{n-1}B_n$        (total permuted ways: n choose n-1)
n+1) $B_1B_2B_3B_4.......B_{n-1}B_n$      (total permuted ways: n choose n)
Also each of the above combinations can be permuted among themself.
The probability of getting the white ball in n+1 th drawing by 1) will be $\binom{n}{0} *\frac{a}{(a+b)}$ by 2) will $\binom{n}{1} * \frac{(a+1)}{(a+b)}$ and so on.............. by n+1) will be $\binom{n}{n} * \frac{(a+n)}{(a+b)}$
Thus the required probability should be $\sum_{i=0}^{n} \binom{n}{i} \frac{a + i}{a + b}$ (since each of the ways are independent of each other)
On solving the above expression we get $$\frac{a * (\sum_{i=0}^{n} \binom{n}{i})  + \sum_{i=0}^{n} i * \binom{n}{i} }{(a+b)}$$
which simplifies to $$\frac{a*2^n + n*2^{n-1}}{(a+b)}$$
