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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!

MathFail
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Sanyam Jain
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2 Answers2

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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: $m+n=\frac{1-b}b, ~m=\frac{a}b$

$$\begin{align}P(X_k=B)&=a\cdot\frac{1-b^{k-1}}{1-b}+b^{k-1}\cdot \frac{a}{1-b}\\ \\ &=\frac{a}{1-b}\\ \\ &=\frac{m}{m+n}\end{align}$$

MathFail
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  • An easy way to see that this can't be right is to consider the case $m=n$. In that case, the answer to the original problem is clearly $\frac 12$ by symmetry. But your formula does not yield $\frac 12$, even when $n=2$. Another way to see it is to note that your expression goes to $0$ as $k\to \infty$, but that doesn't make sense. – lulu Apr 30 '23 at 13:02
  • I didn't get your point, if black balls equal the white balls, everytime you put a white ball to the next bag, so there is one more white ball than black balls in the next bag. – MathFail Apr 30 '23 at 13:10
  • The rules say that you select a ball randomly from an urn and place your selection into the next one. It was not stated that you are only allowed to choose white ones. – lulu Apr 30 '23 at 13:12
  • The OP says "until" so I intepret it as until the k-th drawn is a black ball, then the game stops. – MathFail Apr 30 '23 at 13:15
  • Nothing in the problem statement suggests this. And if this were the case, the answer to the question would clearly be $1$ (as, by definition, you have drawn a black ball from urn $k$). Note that the OP has spelled out a recursion for the desired probability...and that recursion is inconsistent with your interpretation. – lulu Apr 30 '23 at 13:17
  • Why the answer is 1? if $k=1$, the answer should be $m/m+n$ – MathFail Apr 30 '23 at 13:19
  • I thought your interpretation was that you keep playing the game until you draw a black ball from urn $k$. Perhaps you meant something different, but I am quite certain that the OP was not thinking along those lines. As I say, the explicit recursion is not consistent with this reading. – lulu Apr 30 '23 at 13:24
  • I have edited it, without assuming until the black ball was first drawn. – MathFail Apr 30 '23 at 13:53
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    Does that work for the obvious special cases? $m=n$, $n=0$, that sort of thing. In any case, there's a very simple closed formula for the solution. If the OP were to work out even a single case (like $P_2$) I think they would instantly spot the answer, which is then easy to prove. – lulu Apr 30 '23 at 13:57
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    I get the simplified result as $m/(m+n)$ – MathFail Apr 30 '23 at 14:19
  • And that's correct. Easily shown by induction, if nothing else. – lulu Apr 30 '23 at 14:21
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The simplest way is to visualize it as a mixture of black and white balls

and since each bag has an identical mixture at start, on an average, the proportion of black:white will remain unaltered after transfer(s).

Thus P[black will be drawn after transfer(s))] will remain constant at $\dfrac{m}{m+n}$

PS

To reinforce the usefulness of such an approach in simplifying the solution, you might like to have a look at this answer of mine where a more complex probability question has been solved using the "mixture" concept

true blue anil
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