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An urn contains $w$ white and $b$ black balls. $n$ extractions without replacement are made.. Let $I_i$ be a random variable such that $$I_i = \begin{cases} 0 & \text{if the $i$th ball drawn is black,}\\ 1 & \text{if the $i$th ball drawn is white.} \end{cases}$$ and $X_i$ is a random variable that rappresents the total number of white balls extracted during the entire time we are extracting the first $i$ balls $$X_i = \sum_{k=1}^i I_k$$

I know that $$\Pr(X_i=s)=\frac{\dbinom{w}{s}\dbinom{b}{i-s}}{\dbinom{w+b}{i}}$$

but I wondered if it was possible to calculate the same probability via the indicator function defined before.

The single probability is $\Pr(I_i)=\frac{w}{w+b}$.

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No; because the indicator random variables are not independent when drawing samples without replacement.   This isn't an issue with expectations (due to Linearity), but clearly is with probabilities.

You have to consider conditioning, which... works but doesn't lead to a simpler process.

Consider a simple case: $$\begin{align}\mathsf P(I_1+I_2=x) =&~ \sum_{k\in \{0, 1\}} \mathsf P(I_1=k, I_2=x-k) \\[1ex] = &~\mathsf P(I_1=0)\mathsf P(I_2=x\mid I_1=0)+\mathsf P(I_1=0)\mathsf P(I_2=x-1\mid I_1=1) \\[2ex] =&~ \begin{cases}\frac{b}{w+b}\frac{b-1}{w+b-1}+\frac{w}{w+b}\frac{0}{w+b-1} & : x=0 \\ \frac{b}{w+b}\frac{w}{w+b-1}+\frac{w}{w+b}\frac{b}{w+b-1} & : x=1 \\ \frac{b}{w+b}\frac{0}{w+b-1}+\frac{w}{w+b}\frac{w-1}{w+b-1} & : x=2 \\ 0 & : \text{otherwise} \end{cases} \\[1ex] =&~ \dfrac{\binom{w}{x}\binom{b}{2-x}}{\binom{b+w}{2}}\quad\big[x\in \{0,1,2\}\big] \end{align}$$

Then you can use proof by induction to show $$\mathsf P(\sum_{i=1}^n I_i = x) = \dfrac{\binom{w}{x}\binom{b}{n-x}}{\binom{b+w}{2}}\quad\big[x\in \{\max(0,n-b), .., \min(n,w)\}\big]$$

But, way easier to use a combinatorial argument.

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  • $\begingroup$ I've understand, thanks! $\endgroup$ – Paul Jun 15 '16 at 5:42

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