Can a sample space with well defined elements be assigned for every probability experiment?

Question

Can a sample space with well defined elements be assigned for every probability experiment, considering the definition of sample space as-'the set of all distinct elementary events which are all equally likely to occur upon performance of the experiment' ?

Answer 1

I guess I understand now where your confusion comes from. First, we are clear that the answer to your other question is indeed $\frac {49}{80}$ - for all others, please have a look to the comment section.

Now you ask, whether this implies that there have to exist $80$ distinguishable outcomes, the answer is no. Clearly there are only four possible options $$ \left\{ \{{bag}=1,ball=white\},\{{bag}=2,ball=white\},\{{bag}=1,ball=black\},\{{bag}=2,ball=black\} \right\} $$ Important here is, that those events have not the same probability to occur! You can still compute the probability by experimenting and repeat the experiment of pulling balls independently very often, lets say $n(E)$ times, and you count your success of white balls, call it $n(W)$, and then the probability that you pull a white ball is $$ \frac{49}{80}=\mathbf{P}(\text{ball is white})\approx \frac{n(W)}{n(E)} \text{, and equals in the limit} $$

Of course one could also imagine a machine which gives you simply a ball after you started it - in this case you could now imagine that it has one big bag with $49$ white balls and $31$ black balls. In this case you would also have only two possible outcomes (ball is black or white).

When it comes to your definition of sample space as

'the set of all distinct elementary events which are all equally likely to occur upon performance of the experiment'

I am not quite sure what this is supposed to mean. You have a set of possible events which get addressed a probability by performing the experiment. But the term are all equally likely to occur seems strange to me since there is no likelihood we could address those events until we performed the experiment. Otherwise it would be sort of a circular definition.

Maybe this is just supposed to mean that each possible event is only to be allowed once in the set of all events $\Omega$ and not multiple times.

Answer 2

If you are asking whether you can divide up the possible outcomes of a probabilistic event into equally probable outcomes (some of which are identical), then if you accept the existence of $\sqrt{2}$ then certainly not, because an event that turns out "1" with probability $\frac{1}{1+\sqrt{2}}$ and "0" otherwise cannot be expressed using finitely many equally probable outcomes, since the ratio of probabilities of outcomes "1" and "0" is irrational.