(c) $\dfrac{3}{4}$
Sample space = {HH, HT, TH, TT}; total outcomes = 4. Outcomes with at least one tail = {HT, TH, TT} = 3. P(at least one tail) = $\dfrac{3}{4}$.
"At least one tail" means one or more tails, so exclude only the all-heads outcome (HH). The complement method also works: P(no tail) = 1/4, so P(at least one tail) = 1 − 1/4 = 3/4. This mirrors Example 9 of the textbook (at least one head → 3/4).