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Part 1

We could define

"tossing $1$ coin $1$ time" as an experiment $E_1$, with possible outcomes, aka, the sample space of which is $\{H,T\}$, this set has 2 elements;

"tossing $1$ coins $2$ times" as another experiment $E_2$, the sample space of which is $\{(H,H),(H,T),(T,H),(T,T)\}$, this set has 4 elements;

"tossing $2$ coins $1$ time" as third different experiment $E_3$, the sample space of which is $\{(H,H),(H,T),(T,H),(T,T)\}$, this set has 4 elements;

$E_2$ and $E_3$ are two different experiments, though they have same sample spaces.

Part 2

based on the definition of $E_1$, "tossing $1$ coins $2$ times" could also be viewed as "$2$ repetitions" or "carrying out $2$ times" of $E_1$. in this context, the sample space is $\{H,T\}$ rather than $\{(H,H),(H,T),(T,H),(T,T)\}$

Is my understanding above correct?

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  • $\begingroup$ I didn't intervene in your definition of $E_1$, but commented in my answer. You'd better correct it. $\endgroup$ – gunes Feb 28 '19 at 5:15
  • $\begingroup$ What is the "understanding" to which you refer? What exactly are you asking respondents to give opinions about? $\endgroup$ – whuber Feb 28 '19 at 15:05
  • $\begingroup$ Usually the term "outcome" refers to an element of the sample space. You seem to be using the term to refer to the number of coins instead, which is not the usual way the terminology is used. This means that you are miscounting the number of "outcomes" in the usual sense. $\endgroup$ – Ben - Reinstate Monica Mar 4 '19 at 11:28
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Imagine you have a machine that can produce ham. You put a pig (sorry for this pig) in this machine, and you get slices of ham. You can get ham from this machine, but you cannot view this machine as ham.

Now, think of this machine as a function, say y = f(x) = x^2 you can put a value in this function, x = 3 for instance, and then you get y = 9. you can get a value from a function, but you cannot view this function as a value.

random variable is NOT a variable, it is a function

Page 72 in the book “Introduction to Probability, 2nd Edition” says: a random variable is a real-value function of the experimental outcome

based on the above, if you view "tossing 1 coins 2 times" as "2 repetitions" of E_1, you are really tossing the coin in the real world, what you get is value, not a function. so, the sample space is {𝐻,𝑇} for E_1.

if you are considering the set {(𝐻,𝐻),(𝐻,𝑇),(𝑇,𝐻),(𝑇,𝑇)} you are viewing "2 repetitions" as another Experiment that has a sample space equal to this set.

in conclusion, your Part_1 is clear, your Part_2 is answerd by the paragraph above.

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You've a typo in the definition of your first experiment, $E_1$. It should be tossing 1 coin 1 time. If corrected like this, the sample space is correct. Samples spaces of $E_2$ and $E_3$ are also correct, and that they are different experiments having the same sample space. If converted into $\{0,1\}$, you can find many other experiments having the same sample space, e.g. two different match scores (win/lose only) of two footballs teams.

For part 2, carrying out $E_1$ two times creates a different event having a different sample space than $E_1$'s. The elements in tuples can be $H,T$ but this doesn't mean the sample space is $\{H,T\}$. Many events can be decomposed into smaller or unit events having smaller sample spaces.

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  • $\begingroup$ Thanks for the reminder. the example of two footballs teams makes lots of sense. the repetition in Part 2 is used to justify a theoretical value, 50% H, for instance. according to "the law of large numbers", a huge number of repetition would produce a close estimation of true value. So, If I carry out $E_1$ billions of trillions of times. I would get nearly 50% head, in this context, the sample space is {H, T}, the probability of H is 50%. correct me please if something goes wrong. $\endgroup$ – shi95 Feb 28 '19 at 6:16
  • $\begingroup$ Still, repetition of $E_1$ billions of times is another event, say $E_B$, and it has a sample space of tuples like $\{(H,T,...), (T, ...), ...\}$ (having a total of $2^B$ elements). You can devise various statistics from $E_B$ to justify that $P(H)$ is indeed $\frac{1}{2}$ with some confidence, but that's not related to the sample space of $E_B$. $\endgroup$ – gunes Feb 28 '19 at 6:24

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