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We draw (and don't put them back) balls in a box that has $r$ red balls, $y$ yellow balls, $g$ green balls, $b$ blue balls and $w$ white balls. The game stop when we took twice balls of the same color. We are interested on the even

A : "The game stop after $k$ drawn and we drawn two red balls."

What is the probability space that describe this experiment s.t. all elementaries event has same probability ?


Attempts

Q1) What do they mean by "elementary event" ?

Now, I would say that $$\Omega =\{(x_1,...,x_{k-2},R,R)\mid x_i\neq x_{i+1}\text{ for }0\leq i\leq k-3\},$$ but I don't really consider the number of each ball.

Q2) So how can I do here ?

Also, for the event space, we always have $2^{\Omega }$, but the teacher says that here it's wrong but I don't understand why.

Q3) So what's the event space ?

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Let the elements of $\Omega$ be tuples of length $n:=r+y+g+b+w$ such that every tuple contains $r$ numbers $1$, $y$ numbers $2$, $g$ numbers $3$, $b$ numbers $4$ and $w$ numbers $5$.

For the $\sigma$-algebra you can take $2^{\Omega}$.

Then $|\Omega|=\frac{n!}{r!y!g!b!w!}$ and for every $\omega\in\Omega$ we have $$P(\{\omega\})=\frac1{|\Omega|}$$

The event $A$ can be described by:$$\omega\in A\iff|\{\omega_1,\dots,\omega_{k-1}\}|=k-1\text{ and }\omega_{k-1}=\omega_k=1$$


If the $\sigma$-algebra is $2^{\Omega}$ then elementary events are events of the form $\{\omega\}$ where $\omega\in\Omega$.

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  • $\begingroup$ Thank you for your good answer. But my teacher said that : we can't take $2^\Omega $ for the events since there are event that are not observable (but it can be prolonged in $2^\Omega $ in a certain way). For you it doesn't look to be the case... May be you could argue on it ? $\endgroup$ – user601023 Oct 23 '18 at 10:07
  • $\begingroup$ Maybe your teacher was thinking of tuples $(5,12,7,\dots)$ where $\omega_i$ denotes a number of the ball not observable (we only see colors). In that case two distinct tuples can give the same result if you only look at the colors. In my setup we are only dealing with arrangements of colors. $\endgroup$ – drhab Oct 23 '18 at 10:17
  • $\begingroup$ tuple (5,12,7,...) ? What is that ? Indeed, there is something with the fact that we can not distingue two ball of the same color. But indeed, have for example (with your notation), $(1,1)$ and $(1,1)$ is the same... so we should say that $\mathbb P(1,1)=1/\Omega $ since there are several configuration, right ? $\endgroup$ – user601023 Oct 23 '18 at 10:22
  • $\begingroup$ As I said: you can choose for labeling the balls by $1,2,3,\dots,n$ and work in another $\Omega$ where every element $\omega$ is an $n$-tuple such that $\{\omega_1,\dots,\omega_n\}=\{1,2,\dots,n\}$. Then e.g. $\{(5,12,7,\dots)\}$ is the event that first ball $5$ is drawn, then ball $12$, then $7$ et cetera. However, actually the numbers are not observable (only the colors are). Also you could ask your teacher to have a look at his. Actually as long as we are dealing with discrete random variables we can always do it with $2^{\Omega}$ so I don't quite follow your teacher. $\endgroup$ – drhab Oct 23 '18 at 10:30
  • $\begingroup$ Hello, I saw this problem (I had a similar one). Can I ask you a question even if it's not my post ? You are interested on vector of length $n$, i.e. $\omega =(\omega _1,...,\omega _{k-1},\omega _k,...,\omega _n)$ s.t. $\omega _{k-1}=\omega _k=1$ right ? For you $\omega \in A\iff |\{\omega _1,...,\omega _{k-1}\}|=k-1$ and $\omega _k=\omega _{k-1}=1$. But like that, how do you know that there is no $i$ s.t. $\omega _i=\omega _{i-1}$ for $i<k-1$ ? Also, by saying that $|\{\omega _1,...,\omega _{k-1}\}|=k-1$, all $\omega _i$ are different. $\endgroup$ – sam Nov 7 '18 at 13:17

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