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Consider an urn $C$ which contains three (distinguishable) kinds of elements, say $\alpha>0$ elements of kind $A$, $\beta>0$ elements of kind $B$, and $\gamma>0$ elements of kind $G$.

We perform $n>0$ trials (with replacement) of one element at a time from the urn $C$.

We define the event $L^{AB}_n$: "to get at least one element of kind $A$ and at least one element of kind $G$, in $n$ trials".

The probability to get a success for this event in correspondence of the trial $k\in\{1,2\ldots n\}$ is

$$ P(L^{AB}_k)=1-\left(\frac{\alpha+\gamma}{\alpha+\beta+\gamma}\right)^k-\left(\frac{\beta+\gamma}{\alpha+\beta+\gamma}\right)^k+\left(\frac{\gamma}{\alpha+\beta+\gamma}\right)^k. $$

Moreover, it clearly holds the relation

$$ P(\bigcup_{k=1}^n L_{k}^{AB})=P(L_n^{AB}). $$

The union of the events (as expressed above) naturally does not take into account the order in which they occur. But, in our case, it is not possible to get success for the event $L^{AB}_k$ but not for the event $L^{AB}_{k+1}$ (or any following other).

In other words, it seems to me that one should introduce the concept of ordered union of these events.

Is there any definition of such concept?

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  • $\begingroup$ Can you make the urn "G" (or "U", or just call it an urn), and the third kind "C"? $\endgroup$
    – Michael
    Jul 19, 2018 at 7:47
  • $\begingroup$ I don't see a reason to define an ordered union (other than to make indexing easy). The events themselves can specify the "times" they occur, if relevant. $\endgroup$
    – Michael
    Jul 19, 2018 at 7:51

1 Answer 1

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I think the concept you are looking for is a chain of nested events, similar to a chain of nested sets, where each event is an element of the $\sigma$-algebra $\mathcal{F}$ in the probability space you are considering, so each event is a subset of the sample space $\Omega$ and is a subset of the next larger event, i.e. $L^{AB}_{k-1}\subseteq L^{AB}_{k}$

More fully you have $L^{AB}_1 \subseteq L^{AB}_2\subseteq \cdots \subseteq L^{AB}_{k-1}\subseteq L^{AB}_{k}\subseteq \cdots\subseteq L^{AB}_{n}$ so implying $\bigcup\limits_{k=1}^n L_{k}^{AB}=L_n^{AB}$

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  • $\begingroup$ Thanks for your answer, Henry. However, there's no way to take into account (in the union) of the fact that it cannot be $L_k^{AB} \subset L_{k-1}^{AB}$? $\endgroup$
    – user559615
    Aug 24, 2018 at 4:34
  • $\begingroup$ I am not sure about the double negative in your comment, but it is possible to say that (a) the union of a finite chain of nested sets is the largest set, (b) here $L_k^{AB} \nsubseteq L_{k-1}^{AB}$ and $\mathbb P (L_k^{AB}) > \mathbb P (L_{k-1}^{AB})$, and (c) here you could define $L_\infty^{AB} = \sup_k L_{k}^{AB} =\bigcup\limits_{k=1}^\infty L_{k}^{AB}$ and this event has probability $\mathbb P (L_\infty^{AB})= 1$ and so is almost certain $\endgroup$
    – Henry
    Aug 24, 2018 at 8:15

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