Here $S=\{\mathrm{TTT}, \mathrm{TTH}, \mathrm{THT}, \mathrm{HTT}, \mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}, \mathrm{HHH}\}$. Let $E=$ event of getting at most two heads. Then, $E=\{\mathrm{TTT}, \mathrm{TTH}, \mathrm{THT}, \mathrm{HTT}, \mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}\}$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{7}{8} .$$
Explanation as extracted from the printed page; notation may be imperfect.