Solution:
\[\left( iii \right)\] E = event of getting both heads or both tails \[=\text{ }\left\{ HH,\text{ }TT \right\}\]
\[n\left( E \right)\text{ }=\text{ }2\]
Hence, probability of getting both heads or both tails \[=~n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }2/4\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]