Probability dice role
- Thread starter Juju01
- Start date
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,106
Look up the definition you were given for a probability function. You need to determine whether Q(A) satisfies all the requirements.
The main issue will be whether the sum of Q(A) over all possible events A is 1. This will be the sum of all sums, over b. Make b equal to that sum, and you will have the answer.
Incidentally, I don't think I've ever seen the word "eyes" used this way, evidently to mean what I would call "pips".
CORRECTION: Now that I reread this, I'm not sure I understand it. How can a single roll of one die result in an event {1,2,3}? Or, am I confusing events with outcomes?
Last edited:
BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,182
I have the same problem distinguishing between events and outcomes. I think the question misused the word "event A", instead of "outcome A".
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
The way that I read is this: the fact that the die is unfair has no effect on the question unless more information is given.
We are told that the set of outcomes is [imath]~\large\Omega=\{1,2,3,4,5,6\}~[/imath]. Using the standard vocabulary of probability an event is any subset of [imath]~\large\Omega~[/imath]. That is why it is given that [imath]\large sum(\emptyset)=0[/imath]
If that reading is correct the we want to find a [imath]\bf b[/imath] such that [imath]\Large\sum\limits_{A \subseteq \Omega } {Q(A)} = 1[/imath]
[imath][/imath][imath][/imath]
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,106
A common problem in answering questions is that we don't know the context, which is why I often ask for the definitions the student was given, though they rarely respond. I didn't have time before to look up the definition I belatedly realized was in view, so I could check my thinking.
In this case, the "probability function" is not a function of outcomes, but of events; so I think we're talking about something at least close to this:
en.wikipedia.org
In this formulation, the fact corresponding to the sum of probabilities of outcomes being 1 is [imath]Q(\Omega) = 1[/imath] (the second axiom).
Assuming the third axiom (additivity) holds, which we haven't been told to prove, that is equivalent not to [imath]\sum\limits_{A \subseteq \Omega } {Q(A)} = 1[/imath] but to [imath]\sum\limits_{x \in \Omega } {Q(\{x\})} = 1[/imath].
This will also eventually reveal in what way the die is unfair, as one realizes what Q means. (What is Q({1}), for example?)
In this case, the "probability function" is not a function of outcomes, but of events; so I think we're talking about something at least close to this:
Probability axioms - Wikipedia
In this formulation, the fact corresponding to the sum of probabilities of outcomes being 1 is [imath]Q(\Omega) = 1[/imath] (the second axiom).
Assuming the third axiom (additivity) holds, which we haven't been told to prove, that is equivalent not to [imath]\sum\limits_{A \subseteq \Omega } {Q(A)} = 1[/imath] but to [imath]\sum\limits_{x \in \Omega } {Q(\{x\})} = 1[/imath].
This will also eventually reveal in what way the die is unfair, as one realizes what Q means. (What is Q({1}), for example?)
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
Sorry for my mixing up the notation (I fell back into old habits, Prof. Moore used [imath]A+B[/imath] for [imath]A\cup B[/imath].
We want to find a [imath]\bf b[/imath] such that [imath]Q\left(\bigcup\limits_{A \subseteq \Omega } A\right) = 1[/imath] or so that [imath]Q[/imath] is a probability function.
I make [imath]\large\bf b=21\text{ i.e. } sum(\{1,2,3,4,5,6\})=21[/imath]
[imath][/imath]
We want to find a [imath]\bf b[/imath] such that [imath]Q\left(\bigcup\limits_{A \subseteq \Omega } A\right) = 1[/imath] or so that [imath]Q[/imath] is a probability function.
I make [imath]\large\bf b=21\text{ i.e. } sum(\{1,2,3,4,5,6\})=21[/imath]
[imath][/imath]