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Drawing different size balls from a bag (No replacement)

Let's suppose I have a bag with $N$ balls inside. Each ball has a different size. The volume of the $i$th ball is $V_i$ ($i \in [1,N] \land j \in \mathbb{N}$). Let's assume that the probability of drawing the ball $i$ in a turn is proportional to $V_i$. That is to say: The bigger the ball the more likely it will be picked up from the bag. I want to draw $k$ balls from the bag-where $k < N$-one at a time and without replacement of the balls extracted.

Question: What is the probability of drawing the ball $i$ in this process?

This probability has to be expressed in terms of $k$ and $V_j$, where $j \in [1,N] \land j \in \mathbb{N}$.

Thanks!!!

Arnaud Mortier
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user3455372
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  • That's such a fun question, I can't believe you'd rather have it solved by someone else. Anyway, where are you stuck? – Arnaud Mortier Nov 05 '19 at 20:19
  • First you should start with: Given the subset of balls $I\subseteq {1,2,3,\dots,n}$ the probability of $i\in S$ being drawn is $\frac{V_i}{\sum_{i\in S} V_i}.$ – Thomas Andrews Nov 05 '19 at 20:20
  • (I'm not sure if there is a "nice" answer.) – Thomas Andrews Nov 05 '19 at 20:21
  • This is called "weighted random sampling without replacement" or similar. There are various resources if you google, including several related threads here on MSE. In short, unlike the unweighted case, there is no really simple solution, but there are some techniques available. – antkam Nov 05 '19 at 22:08

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