There are $k ≥ 3$ bags, each containing m black and n white marbles. A marble is randomly chosen from
the first bag and transferred to the second bag; then a marble is randomly chosen from the second bag and transferred to the third bag, and so on until a marble is randomly chosen from the kth bag.
Find the probability that the marble chosen from (i) the 3rd bag is black; (ii) the $k$-th bag is black.
I don't understand how to proceed in this. I thought of trying recursion but it doesn't seem to be working. For the 3 bags case I think it is easy to solve by individually considering each case but for the kth part I have zero idea. Kindly help.
The way I tried recursion:
Suppose $P_k$ is the probability of drawing black ball from $k$-th bag.
Then,
$$P_k=P_k-1*(m+1)/(n+m+1)+(1-P_k-1)*m/(n+m+1)$$
I don't see a way of proceeding further, Plus this seems to be a bit too complicated. I was wondering if there is a simpler solution.
Thank you!
Best Answer
Let $X_k$ be the k-th draw, and $B$ for black ball, $W$ for white ball.
$$\begin{align}P(X_3=B)&=P(X_3=B\cap X_2=W)+P(X_3=B\cap X_2=B)\\ \\ &=P(X_3=B|X_2=W)\cdot P(X_2=W)+P(X_3=B|X_2=B)\cdot P(X_2=B)\\ \\ &=\frac{m}{m+n+1}\cdot P(X_2=W)+\frac{m+1}{m+n+1}\cdot P(X_2=B)\\ \\ &=\frac{m}{m+n+1}\cdot (1-P(X_2=B))+\frac{m+1}{m+n+1}\cdot P(X_2=B)\\ \\ &=\frac{m}{m+n+1}+\frac{1}{m+n+1}\cdot P(X_2=B) \end{align}$$
with initial condition,
$$P(X_1=B)=\frac{m}{m+n}$$
For the general k-th draw, we have the recursion equation:
$$\begin{align}P(X_k=B)&=\frac{m}{m+n+1}+\frac{1}{m+n+1}\cdot P(X_{k-1}=B)\\ \\ P(X_k=B)&=a+b\cdot P(X_{k-1}=B)\end{align}$$
where,
$$a=\frac{m}{m+n+1},~~ b=\frac{1}{m+n+1}$$
Solve this recursion equation and we get:
$$\begin{align}P(X_k=B)&=a+ab+ab^2+...+ab^{k-2}+b^{k-1}\cdot P(X_1=B)\\ \\ &=a\cdot\frac{1-b^{k-1}}{1-b}+b^{k-1}\cdot \frac{m}{m+n}\end{align}$$
Note that: $m+n=\frac{1-b}b, ~m=\frac{a}b$
$$\begin{align}P(X_k=B)&=a\cdot\frac{1-b^{k-1}}{1-b}+b^{k-1}\cdot \frac{a}{1-b}\\ \\ &=\frac{a}{1-b}\\ \\ &=\frac{m}{m+n}\end{align}$$