How do we prove that the probability remains the same?

Question

From an urn containing $M$ white balls and $N-M$ black balls, $r$ balls have been lost. The probability of drawing a white ball is: Given ($N\gt M\gt r)$.

My Thinking: We know that before the balls have been lost $P$(white)=$\frac{M}{N}$. Now when $r=1$, $$P_{white}=(\frac{M}{N})(\frac{M-1}{N-1})+(\frac{N-M}{N})(\frac{M}{N-1})=\frac{M}{N}$$ but now for $r=2$ ($3$ possibilities), the situation gets bit complicated and further for $r=3$ gets further tough. So I want to derive(I think it will involve binomial maybe) a general term for this probability and also to prove that the probability always remains i.e, $P$(white)=$\frac{M}{N}$ independent of the balls drawn. How can we prove this?

Answer 1

The probability that a particular while ball is drawn is the probability that it was not lost, times the conditional probability that it is drawn, given that it wasn't lost.

The probability that it wasn't lost is $$\frac{N-r}{N}$$ and the conditional probability that is drawn, given that it wasn't lost is $$\frac1{N-r}$$ so that probability that it is drawn is $$\frac{N-r}{N}\cdot\frac1{N-r}=\frac1N$$

Summing over all white balls, we find the probability that a white ball is drawn is $$\frac MN$$

Of course, I'm assuming that each ball has an equal probability of being lost, and that all relevant events are independent.

Answer 2

I regard this is an intuition question rather than a math question.

Consider the following two scenarios:

• Scenario 1: First you choose a ball at random. Then, from the remaining $(N-1)$ balls, you set $(r)$ balls aside.

• Scenario 2: First, you set $(r)$ balls aside, chosen at random. Then, from the remaining $(N-r)$ balls, you choose one at random.

Clearly, in scenario 1, the chance that the (first) ball chosen is white is unaffected by your having subsequently set aside $(r)$ balls. Therefore, the challenge is to demonstrate, by analysis, that the chance of a white ball chosen in scenario 2 is identical to the chance of a white ball chosen in scenario 1.

I am assuming that the $(r)$ balls set aside are always distinct from the single ball chosen to see if the single ball is white or black.

Vary scenario 2 into scenario 3:
Instead of choosing $(r)$ balls at random, you choose $(r + 1)$ balls at random. Then, from the $(r + 1)$ balls, you randomly select one and examine whether that ball is white or black.

Is this equivalent to scenario 2?

In scenario 2, you first chose $(r)$ balls and set them aside. Then, you chose a ball at random from the remaining $(N-r)$ balls. Before any selections were made, each of the $(N)$ balls was just as likely to be white as any other ball. Therefore, choosing a ball at random from the $(r+1)$ balls is just as likely to produce a white ball as choosing a ball at random from the $(N-r)$ balls. That is, since the $(r)$ balls being set aside are chosen at random, none of them is more or less likely to be white than any of the $(N-r)$ balls that remain.

This means that the chance of a white ball is the same, in scenario 3 as scenario 2.

What about scenario 3 versus scenario 1?

In scenario 3, you first choose $(r + 1)$ balls, and then choose one at random from them. As discussed, since the $(r + 1)$ balls are chosen at random, the chance of any one of them being white is the same as the chance of any of the $(N)$ balls being white. Therefore, with respect to the chance of choosing a white ball, scenario 3 is equivalent to scenario 1.

Answer 3

Assuming we have w white balls and b black balls, define the following event:
$A_{w,b}^{r}$:= after losing r balls from an urn containing w white and b black balls, we choose a ball at random and it is white.

Now by induction on $r$, assuming that for all $r'<r, \mathcal{P}(A_{w,b}^{r'})=\frac{w}{w+b}$, we write:

$\begin{equation} \begin{split} \mathcal{P}(A_{w,b}^{r})&= \mathcal{P}(A_{w,b}^{r}|\text{first lost ball was white})\cdot\mathcal{P}(\text{first lost ball was white}) + \mathcal{P}(A_{w,b}^{r}|\text{first lost ball was black})\cdot\mathcal{P}(\text{first lost ball was black})\\&=\mathcal{P}(A^{r-1}_{w-1,b}) \cdot\frac{w}{w+b} + \mathcal{P}(A^{r-1}_{w,b-1}) \cdot\frac{w}{w+b}\\&=\frac{w-1}{w+b-1}\cdot\frac{w}{w+b}+\frac{w}{w+b-1}\frac{b}{w+b}\\&=\frac{w\cdot(w-1+b)}{(w+b-1)\cdot(w+b)}\\&=\frac{w}{w+b} \end{split}\end{equation}$

Of course the base of induction is true for r=1.

Answer 4

First note that $\binom{n}{k}(n-k)=n\binom{n-1}{k}$, which we use twice below. (This binomial coefficient identity can be proved combinatorially by counting, from among $n$ people, committees of size $k+1$ that include a chairperson. The LHS chooses everybody but the chair and then chooses the chair from the remaining $n-k$ people. The RHS chooses the chair and then chooses the remaining $k$ members.)

By conditioning on the number $w$ of white balls lost, we obtain that the probability of white is $$\sum_{w=0}^r \frac{\binom{M}{w}\binom{N-M}{r-w}}{\binom{N}{r}}\cdot \frac{M-w}{N-r} =\frac{M}{N}\sum_{w=0}^r \frac{\binom{M-1}{w}\binom{N-M}{r-w}}{\binom{N-1}{r}} =\frac{M}{N},$$ where the last equality follows from the hypergeometric distribution or via Vandermonde's identity.