Consider tossing a fair coin once followed by rolling a fair die, however the die is rolled once only when we get a head in the preceding toss. Need to find out probability of getting a six or a tail. I could derive the answer rather mechanically using a tree diagram. But I'm looking for a more convincing solution perhaps using set notation and formula. Any help would be much appreciated.
Elaboration of my attempt and confusion:
Sample space $\Omega=\{\{H,1\}, \{H,2\}, \{H,3\}, \{H,4\}, \{H,5\}, \{H,6\}, \{T\}\} $ [note that all sample points are not equi-dimensional]
Events are defined for the same sample space as follows:
$A=$the event that we get a H in the toss, i.e. all 6 points starting with H
$B=$the event that we get a six in the roll of a die i.e. a single point ending with 6
$C=$the event that we get a T in the toss i.e. a single point with T
$ℙ$(getting a six or a tail) = $ℙ(B∪C) = ℙ(B) + ℙ(C) - ℙ(B∩C) = ℙ(B) + ℙ(C) – 0$
However, at this point, I’m not sure how to proceed further to calculate $ℙ(B)$?
One, thought could be replacing $B$ with $\{H,6\}$ i.e. $ℙ(B) = ℙ(\{H\}∩\{6\})$. However, here I’ve two confusions –
(1) If I use events like $\{H\}$ and $\{6\}$, then they are scoped within respective sub-experiments of tossing and rolling. Moreover, such decomposition is allowed only-when the final space $\Omega$ can be expressed as Cartesian product space $S_1\times S_2$ where $S_1=$ sample space of tossing and $S_2=$sample space for rolling
(2) How can I define another event say $B^1$ in $\Omega$, such that $B=\{A \cap B^1 \}$