JEE Advanced 2021 Paper 2 Mathematics: Paragraph 2

Let $E$ be the expected number of tosses from the start and $E_H$ the expected further tosses given the last toss was a head (and no $HH$ yet). From the start: $E=1+\frac12E_H+\frac12E$. From state $H$: $E_H=1+\frac12\cdot0+\frac12E$. Substituting, $E=1+\frac12\left(1+\frac E2\right)+\frac E2$, so $E=\frac32+\frac{3E}{4}$, hence $E=6$. This agrees with summing $\sum nP(X=n)$ using $P(X=n)=F_{n-1}/2^n$ (Fibonacci numbers), which I confirmed by exact numerical summation. So $E(X)=6$.

Conditioning on the first tosses: $X=n$ requires the sequence to avoid $HH$ and end with $THH$ (for $n\ge3$), so $P(X=n)=F_{n-1}/2^n$ with $F_1=F_2=1$; thus $P(X=2)=\frac14$, $P(X=3)=\frac18$, $P(X=4)=\frac18$, $P(X=5)=\frac3{32}$, and so on. Use states: let $u$ be the probability that the number of remaining tosses is even when the last toss was not a head (or at the start), and $v$ the same probability when the last toss was a head. Then $v=\frac12(1-u)$, because a head stops the game after one (odd) toss, while a tail returns to the first state after one toss; and $u=\frac12(1-u)+\frac12(1-v)$. Substituting $v$ gives $u=1-\frac u2-\frac{1-u}4=\frac34-\frac u4$, so $\frac54u=\frac34$ and $u=\frac35$. Direct numerical summation of the series to 400 terms agrees, so the probability is $0.60$.