Answer

Let \(X\) be the # of tosses till first H. Then, \((X = 1) \cap (X > 1) = \phi\). Using Total Expectation Theorem

\[ \begin{aligned} E[X] &= P(X = 1)E[X|X = 1] + P(X > 1)E[X|X > 1] \newline &= 0.5 * 1 + 0.5 E[X] \newline \Rightarrow E[X] &= 2 \end{aligned} \]

\(P(X = 1) = 0.5\) because then we get the head in the first toss itself. Since \(P(X = 1) + P(X > 1) = 1\), we have \(P(X > 1) = 0.5\). \(E[X] = E[X|X > 1]\) because the tosses are independent and thus memoryless.