Double coin toss

Double coin toss. We wanna know what’s the probability of throwing at least one head.

  1. First we model it as $Omega$ = {{H $cap$ H}, {H $cap$ T}, {T $cap$ H}, {T $cap$ T}}. I often see here that people solve: P({H $cap$ H} $cup$ {H $cap$ T} $cup$ {T $cap$ H}) = P(H $cap$ H) + P(H $cap$ T) + P(T $cap$ H). But this is not possible because Kolmogorov’s axiom #3 just allows addition of disjunct events/sets, but (H $cap$ T) and (T $cap$ H) are not disjunct. So in this model we’d need a different way to answer the question, e.g. thru using the probability of the complement.

  2. Now we model it as $Omega$ = {{H1 $cap$ H2}, {H1 $cap$ T2}, {T1 $cap$ H2}, {T1 $cap$ T2}}. Now we can apply the addition rule: P({H1 $cap$ H2} $cup$ {H1 $cap$ T2} $cup$ {T1 $cap$ H2}) = P(H1 $cap$ H2) + P(H1 $cap$ T2) + P(T1 $cap$ H2) because all events/sets are disjunctive.

  3. In model 1, Bayes’ theorem holds, because (H $cap$ T) = (T $cap$ H) by commutativity and therefore its probabilites are the same which is the condition to derive Bayes’ theorem. But in model 2, Bayes’ theorem does not hold because you do not have a relation of commutativity like in 1, but you need it, else it’s not a theorem there.

Can you tell me if I am right in all three points, and if not why?