= Two-toss sample space
{title2=$\Omega=\{HH,HT,TH,TT\}$}
Model two fair <independent> coin tosses, recording their order. Here $H$ denotes heads and $T$ tails, and
$$
\Omega=\{HH,HT,TH,TT\},\qquad
\mathcal F=\mathcal P(\Omega),\qquad
\mathbb P(\{\omega\})=\frac14\quad(\omega\in\Omega).
$$
The <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. Then
$$
A=\{HH,HT\},\qquad B=\{HH,TH\},\qquad
A\cup B=\{HH,HT,TH\},\qquad A\cap B=\{HH\}.
$$
The <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> $1/4$.
\Image[/two-coin-outcomes-and-events.png]
{title=Four ordered outcomes and the event of at least one head}
{description=Rows record the first toss and columns the second. Each cell has <probability> $1/4$ under fair <independent> tosses. The highlighted <union> $A\cup B$ includes $HH$, $HT$, and $TH$; the overlap $A\cap B$ contains only $HH$.}
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 $\mathbb P(A\cap B)=3/8\ne1/4$, 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.
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