I'm trying to find the probability of an outcome where, using the traditional example, white balls are replaced by black balls once selected.

Initially I have $n$ white balls and $\mu$ samples. I want to find the probability of selecting $b$ of these, whereby after selecting a white ball it is replaced by a black ball.

To complicate the matter, there is a probability $1-\eta$ per sample that I won't successfully select a white ball ($\eta$ representing the chance of successfully picking a ball), even if the bag is filled only filled with white balls. This $\eta$ is uniform across all $n$ balls.

By sitting and analyzing the probability trees, I have devised the following equation, but I believe this can be entirely rewritten by someone more competent with probabilities. The first term (the product) creates the probability of the first $b$ samples all being correct. The second term creates every conceivable combination of failure that can happen and still manage $b$ successfully.

$P_{output}(\mu,b,n)= \left(\prod\limits_{i=0}^{b-1}\eta\frac{n-i}{n}\right)\cdot\left(1+\sum\limits_{j_1=0}^{b-1}\left(1-\eta\frac{n-j_1}{n}\cdot\sum\limits_{j_2=j_1}^{b-1}\left(1-\eta\frac{n-j_2}{n}\cdot \cdots\sum\limits_{j_{\mu-b}=j_{\mu-b-1}}^{b-1}\left(1-\eta\frac{n-j_{\mu-b}}{n}\right)\right)\right)\right)$

Thank you in advance for your help!

Edit: I'm wondering whether this can be represented as an urn problem, where there are $n$ different coloured sub-urns inside this urn. Inside each sub-urn there are $\eta/\eta=1$ white balls, and $(1-\eta)/\eta$ black balls. So, you need the probability of selecting $b$ different sub-urns multiplied by the probability of successfully selecting the one white ball from that urn, for $\mu$ samples. Does that work?


As per the answer you provided, you can express this using Markov chains, with states $0$ to $n$, and state $i$ indicating you've already put $i$ black balls into the urn. The chance of moving from state $i$ to state $j$ after a single selection is then the $i^{\textrm{th}}$ row and $j^{\textrm{th}}$ column of

$P = \begin{pmatrix} (1-\eta) & \eta & 0 & 0 & \cdots & 0\\ 0 & (1-\frac{n-i}{n}\eta) & \frac{n-i}{n}\eta & 0 & \cdots & 0 \\ 0 & 0 & (1-\frac{n-i}{n}\eta) & \frac{n-i}{n}\eta &\cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}.$

We can then find the probability of being in state $i$ after $\mu$ samples, from initial state $0$, as the expression

$(1,0,0,\ldots,0) P^{\mu} = l^T p^{\mu}.$

$P^{\mu} l,$ as you suggested, would give you the vector whose $i^{\textrm{th}}$ element is the probability of ending in state $0$ starting from state $i,$ which can easily to be shown to be the vector



After some analysis I've discovered this can be represented as a Markov chain with bidiagonal transition matrix:

$\mathbb{P} = \begin{pmatrix} (1-\eta) & \eta & 0 & 0 & \cdots & 0\\ 0 & (1-\frac{n-i}{n}\eta) & \frac{n-i}{n}\eta & 0 & \cdots & 0 \\ 0 & 0 & (1-\frac{n-i}{n}\eta) & \frac{n-i}{n}\eta &\cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$

where $i$ is the row number (from $i=0$ to $i=b$). By raising this transition matrix to the power $\mu$, and with initial state $l$ (where rows go from $i=0$ to $i=b$):

$l = \begin{pmatrix} 1\\ 0\\ \vdots\\ 0 \end{pmatrix}$

it is possible to find probability as the population of the final row in the equation:

$a = \mathbb{P}^\mu l$


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