A fair coin is tossed 6 times. Let E be the event that no two consecutive tosses are both heads, and F the event that exactly three heads occur.

Count sequences of length $6$ with no two adjacent H: they are counted by Fibonacci numbers, $F_8=21$, so $P(E)=\frac{21}{64}$. Sequences with exactly $3$ heads and none adjacent: choose $3$ non-adjacent positions out of $6$, $\binom{6-3+1}{3}=\binom43=4$, so $P(E\cap F)=\frac4{64}=\frac1{16}$. Also $P(F)=\binom63/64=\frac{20}{64}$. Therefore $P(F\mid E)=\frac{4}{21}$ and $P(E\mid F)=\frac{4}{20}=\frac15$. (A) is correct. (B) $\frac16\ne\frac4{21}$; incorrect. (C) $\frac14\ne\frac15$; incorrect. (D) $\frac18\ne\frac1{16}$; incorrect. Exhaustive enumeration of all $64$ sequences confirms these values. Answer: A only. A common slip is to confuse $P(E\mid F)$ with $P(F\mid E)$: the conditioning event changes the denominator from $20$ to $21$ favourable sequences.