Problem

Solution (clear steps)

Step 1 — Sample space.
For three fair coin tosses the sample space has \(2^{3}=8\) equally likely outcomes: \[ \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}. \]

Step 2 — Describe events A and B.
\[ A=\{\text{exactly 2 heads}\}=\{HHT, HTH, THH\},\quad |A|=3. \] \[ B=\{\text{at most 2 tails}\}=\{\text{outcomes with } \#\text{tails}\le 2\}. \] The only outcome with more than 2 tails is \(TTT\), so \(|B|=8-1=7\) and \[ B=\{HHH, HHT, HTH, THH, HTT, THT, TTH\}. \]

Step 3 — Relation between A and B.
Every outcome in A has exactly one tail, so \(A\subseteq B\). Therefore \[ A\cup B = B. \]

Final answer: \[ P(A\cup B)=P(B)=\frac{|B|}{8}=\frac{7}{8}. \]