I'm a little confused about what this problem is asking exactly and would simply like some advice on my solution. The problem is:
We roll a die three times. Give a probability space $(\Omega, \mathcal F, P)$ for this experiment.
The following is my written solution.
Assuming the die is fair and six-sided, the sample space for one roll is $\Omega_1 = \{1, 2, 3, 4, 5, 6 \}$. Which, for all three rolls gives:
$$ \Omega = \{ (1,1,1), …, (1,6,6), (2,1,1), …, (2,6,6),(3,1,1),…,(5,6,6),…,(6,6,6) \} $$
Our sigma field is then
$$ \mathcal F = \{ \varnothing, \Omega, \{1\}, …,\{6\},\{1,1\},\{1,2\},…,\{5,6\},\{6,6\}\} $$
Now, let $A\subset \Omega$ such that $ A=A_{111}\cup A_{112} \cup … \cup A_{665} \cup A_{666}= \bigcup_{i,j,k=1}^6 A_{ijk}$.
Then, $$ P(A)=\frac{A}{\Omega} $$
Our probability space is then defined by $(\Omega, \mathcal F, P(A))$, whose values are listed above.
Here are my questions regarding this problem:
1) Are my sample space and sigma field correct for this experiment?
2) The initial question was a little vague and am unsure of what $P$ I'm looking for here, so I took a guess at that solution. I'm fairly certain my answer for that part is incorrect. From examples I've seen online, a specific event is typically given, and you're required to find the probability of that event. So, I tried to expand that to include any possible rolling combinations.
To clarify some notation just in case, $A_{111}$ is the event that you roll a 1 three times. Similarly, $A_{352}$ would be the event that you roll a 3, then a 5, then a 2.
Thank you!
Best Answer
I think you have a good hang of the concept. However, things can always be written better.
For example, the sample space for three independent dice, rather than the suggestive $\{(1,1,1),...,(6,6,6)\}$(which is correct, so credit for that) can be written succinctly as $\Xi \times \Xi \times \Xi$, where $\Xi = \{1,2,3,4,5,6\}$. This manages to express every element in the sample space crisply, since we know what elements of cartesian products look like.
The sigma field is a $\sigma$-algebra of subsets of $\Omega$. That is, it is a set of subsets of $\Omega$, which is closed under infinite union and complement. Ideally, the sigma field corresponding to a probability space, is the set of events which can be "measured" relative to the experiment being performed i.e. it is possible to assign a probability to this event, with respect to the experiment being performed. What this specifically means, is that based on your experiment, your $\sigma$-field can possibly be a wise choice.
In our case, we have something nice : every subset of $\Omega$ can be "measured" since $\Omega$ is a finite set (it has $6^3 = 216$ elements) hence the obvious candidate, namely the cardinality of a set can serve as its measure(note : this choice is not unique! You can come up with many such $\mathcal F$).
Therefore, $\mathcal F = \mathcal P(\Omega)$, where $\mathcal P(S)$ for a given set $S$ is the power set of $S$, or the set of all subsets of $S$. This is logical since this contains every subset of $\Omega$, and is obviously a $\sigma$-algebra.
Now, $P(A)$, for $A \subset \mathcal F$(equivalently, $A$ any subset of $\Omega$) is, for the reasons I mentioned above, simply the ratio between its cardinality, and the cardinality of $\Omega$. Therefore, $P(A) = \frac{|A|}{216}$. That is exactly what you wrote, except well, division by sets isn't quite defined.
So there you have it, an answer, along with what you've done right and wrong.
NOTE : $\mathcal P$ for the power set, and $P$ for the probability of a set is my notation here. I still think the letter $P$ is used for both, but it's causing confusion here, hence the change.