# 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?