# 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:$1-($($b/(a+b)$)$(1-1/(a+b))^{n}$)

• Michael already gave you a very good hint. You say that you don't get the correct answer. Please edit your question to show the work that you've done implementing his hint, and what you computed, and what you know is the correct answer. – Willie Wong Nov 22 '13 at 12:21
• Yes,Answer is Updated.. – anshul22 Nov 22 '13 at 12:31
• And how have you used Michael's hint? – Willie Wong Nov 22 '13 at 12:44
• Frankly,speaking i'm assuming that there is lack in the Hint i.e.,what is w? and that w is not there in Answer also!! – anshul22 Nov 22 '13 at 12:52
• $w$ stands for white, the number of white balls, which you call $a$. – Willie Wong Nov 22 '13 at 13:15

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:

1. Given the above: what is the probability that black ball number $l$ is picked on the $k$th draw?
2. Given that there are $b$ black balls, what is the probability that a black ball is picked on the $k$th draw?
3. 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?
4. What is the correct $k$ to use if you have already repeated the procedure $n$ times and are about to pick another ball?

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 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 $$\frac{a*2^n + n*2^{n-1}}{(a+b)}$$

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.

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

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.