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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 two events:

  • $L^{AB}_n$: "to get at least one element of kind $A$ and at least one element of kind $G$, in $n$ trials"
  • $E_n^G$: "to get $n$ elements of kind $G$, in $n$ trials"

I would like to study the probability that one trial, out of $n$, is a successful trial for these events, separately.

My reasoning, based on the definition of probability as the ratio between favorable and possible cases, is: there are $n$ possible trials (e.g. $n$ possible cases), but not all of them represent a favorable case.

In particular:

  • There are $n-1$ trials which represent a favorable case, out of $n$, for the success of the event $L^{AB}_n$, although not all of these possible cases have the same chance to occur. In fact $$ P(L^{AB}_k)=1-\left(\frac{\alpha+\gamma}{\alpha+\beta+\gamma}\right)^k-\left(\frac{\beta+\gamma}{\alpha+\beta+\gamma}\right)^k+\left(\frac{\gamma}{\alpha+\beta+\gamma}\right)^k. $$
  • There is only one trial which represents a favorable case, out of $n$, for the the success of the event $E^{G}_n$, and $$ P(E^{G}_{n,k})=\left(\frac{\gamma}{\alpha+\beta+\gamma}\right)^n\delta_{n,k}, $$ where $P(E^{G}_{n,k})$ is the probability to get a success for $E^{G}_{n}$ at the $k$-th trial, and $\delta_{n,k}$ is the Kroneker symbol.

I point out that I am interested to know the probability that one of the $n$ trials is a success, and not on the probability of the events to occur in the course of the $n$ trials (these probabilities are already calculated above). In other words, I am looking for a probability in the form

$$ \frac{\text{$m$ favorable trials}}{\text{$n$ possible trials}}. $$

I would say that:

  • The probability that one trial is a success for the event $L^{AB}_n$ is $$ \frac{\sum_{k=2}^{n}[P(L^{AB}_k)\cdot 1]}{n}, $$

rather than simply $\frac{n-1}{n}$, because we must weight each of the $n-1$ favorable trials at the numerator with the probability that the event $L^{AB}_n$ occurs in correspondence of each trial.

  • The probability that one trial is a success for the event $E_n^G$ is $$ \frac{\sum_{k=1}^{n}[P(E^{G}_{n,k})\cdot 1]}{n}=\frac{\left(\frac{\gamma}{\alpha+\beta+\gamma}\right)^n}{n}, $$

rather than simply $\frac{1}{n}$, since the probability that the only possible favorable case (i.e. the last trial) is a success is $\left(\frac{\gamma}{\alpha+\beta+\gamma}\right)^n$.

Is this reasoning correct?

I am almost sure that it is not correct! Therefore I need your help to understand the conceptual mistakes that I am doing here. I however apologize for naivety and silly errors.

NOTE: I posted some other question with a similar topic: for instance, Time to first success in case of increasing probability at each trial, Time to first success of an event-intersection, and Time to first success for a simple event.

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  • $\begingroup$ What does it mean "the probability that one trial, out of n, is a successful trial for these events", i.e. what is $L_k^{AB}$? $\endgroup$ – d.k.o. Jul 19 '18 at 7:04
  • $\begingroup$ @d.k.o. This means that, in correspondence of the trial $k$, you get at least one element of kind $A$ and at least one element of kind $B$. Imagine, for instance, that you got an element of kind $A$ in a previous trial: if, at the trial $k$, you get one element of kind $B$, then this trial represents a "successful one", since you got a success for the event $L_n^{AB}$. $\endgroup$ – user559615 Jul 19 '18 at 7:25

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