I have a question: An experiment: a coin is flipped. if we have HEAD than we flip a tetrahedron $(t_1 \ldots t_4)$. if we have Number than we flip a cube $(c_1 \ldots c_6)$.
1.Now i am to find Ω.And here am not sure.
Ω={t1..t4,if H}U{c1..c6 ,if N}=10.
2.And for example ,probability to get odd number of cube flipping is 3/10 or 6/10 or 1/4?
thank you!
First of all, I would write the elements of $\Omega$ as pairs: one side of the coin, followed by the result of the rolling whichever polyhedron you roll next.
For example, one element is $(H, 3).$ If you like you can write it $(H, t_3)$ to remind you that the second part of the pair is a side of a tetrahedron.
There are $10$ elements of $\Omega$ but I think it is not reasonable to assign equal probability to each one unless it is your intention to model a coin that is unfair and lands heads $60\%$ of the time. You can partition $\Omega$ into the subsets with $H$ and the subsets with $T$; if the coin is fair the two subsets should have equal probability measure and you can proceed from there.