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Assume we have red and a black cube (normal cubes with 6 sides). We roll these two dice. Is it true that the sample space is $$\Omega = \left\{ (r,s) \mid r \in \{ 1, \dots, 6\}, s \in \{ 1, \dots, 6\} \right\}?$$

I'm now interested in the event $A =$'The red cube shows an even number' for example. Can I define the set $A$ like
$$A = \{(r,s) | r \in \{2,4,6\}, s \in \{ 1 \dots 6\} \}?$$ How can I define $A$'s probability density function and probability measure for this probability space?

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your event $A$ is defined correctly. Before defining a probability measure you have to define a simga-field. intuitively, the sigma-field contains all sets which you want to be able to assign a probability to. Most often we simply choose the power set of $\Omega$ as the corresponding sigma field $\mathcal F$, i.e. $$ \mathcal F = \mathcal P (\Omega). $$

By doing so we are able to define a probability measure $\Bbb P$ on $\mathcal{F}$ which assigns a probability to every possible event. Since every event has the same probability and there is a total of 36 events we could define: $$ \Bbb P: \mathcal F \rightarrow [0,1]: A \mapsto \frac{|A|}{36} $$ In particular, if we want to compute the probability that the red dice is even we have to count all events where this is the case. It is easy to verify that in 18 ($ = 3*6$) cases the red dice is red. Therefore $$ \Bbb(P)(A) = \frac{18}{36} = \frac 12, $$ in this case.

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  • $\begingroup$ How you know all points have same chance??? he mentioned a cube with 6 side! how you know all side is equal? Match is a cube with 6 side. $\endgroup$ – masoud Mar 26 '19 at 10:01
  • $\begingroup$ Since the OP did not specify the probabilities for each side I assumed that the dice is fair, and hence that each side has the same probability. What makes you think otherwise? $\endgroup$ – Cettt Mar 26 '19 at 10:02
  • $\begingroup$ if the question say all points have same chance , so your way is correct. i do not know what means "normal cube" means??.assume OP say "a cube with 6 side" (nothing more), you can not add another info to problem!! can we add further information to problem? how you know it is correct? you should have a better reason? $\endgroup$ – masoud Mar 26 '19 at 10:25
  • $\begingroup$ I think in this case, $F=\{\phi ,\Omega\}$ and can only calculate $P(\phi )=0$ $P(\Omega)=1)$. an example: let we have one cube, that $D=\{ 1,2\})$ and $B=\{ 3,4,5,6\}$ have same chance. what is the $P(A=\{2,4,6\})$? (note by that info i told $F=\sigma \{D,D^{c} \}$) we can not talk about the events like $A$ that have no information about it. since we can not calculate it $\endgroup$ – masoud Mar 26 '19 at 10:25
  • $\begingroup$ well for me a 'normal' cube is a 'fair' cube, but please feel free to post your answer $\endgroup$ – Cettt Mar 26 '19 at 10:34

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