Find the probability of almost two talks or atleast two heads I a tips of three coins

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$$\begin{gathered} \mathrm{In} \mathrm{a} \mathrm{toss} \mathrm{of} \mathrm{three} \mathrm{coins}: \\ \mathrm{Sample} \mathrm{space}= \\ \left\{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{TTT},\mathrm{TTH},\mathrm{THT},\mathrm{THH}\right\}=8 \\ \mathrm{atmost} \mathrm{two} \mathrm{tails}=\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{TTH},\mathrm{THH},\mathrm{THT},\mathrm{HHH}=7 \\ \mathrm{hence} \\ \mathrm{P}\left(\mathrm{atmost} \mathrm{two} \mathrm{tails}\right)=\frac{7}{8} \\ \mathrm{atleast} \mathrm{two} \mathrm{heads}=\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{THH}=4 \\ \mathrm{P}\left(\mathrm{atleast} \mathrm{two} \mathrm{heads}\right)=\frac{4}{8}=\frac{1}{2} \end{gathered}$$

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