 Two-toss sample space

Model two fair independent coin tosses, recording their order. Here $H$ denotes heads and $T$ tails, andThe power set $\mathcal{P}(\Omega )$ includes every question expressible as a subset of these four outcomes. The mass $1/4=(1/2)(1/2)$ comes from the fairness and independence assumptions. These are exact model probabilities, rather than measured toss frequencies.

Let $A$ be the event that the first toss is heads and $B$ the event that the second toss is heads. ThenThe union means “at least one head,” while the intersection means “both heads.” Their probabilities are $3/4$ and $1/4$. Counting the union through the inclusion-exclusion principle gives $1/2+1/2-1/4=3/4$; alternatively its complement of an event is $\{TT\}$, of probability $1/4$.

On the same sample space, assigning masses $3/8,1/8,1/8,3/8$ to $HH,HT,TH,TT$ would still make each toss fair: $\mathbb{P}(A)=\mathbb{P}(B)=1/2$. But $\mathbb{P}(A\cap B)=3/8\ne 1/4$, so the tosses would be dependent. The event of at least one head would have probability $5/8$. Changing the probability measure changes the answer even when the sample space and the single-toss probabilities stay the same.