Find $i \div j$.

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9. Tossing a coin 10 times has $2^{10}$ outcomes. Let $A(n)$ be the number of outcomes where no heads appear consecutively. By enumeration, it is easy to see that $A(1)=2, A(2)=3, A(3)=5$. It can also be proven by induction that there is a recurrence relation:
$$
A(n+2)=A(n+1) + A(n) \text {. }
$$

This is because $A(n+2)$ includes two types of sequences: one where no heads are consecutive and ends with a tail, of which there are $A(n+1)$; and another where no heads are consecutive but ends with a head, of which there are $A(n)$. Thus, $A(n)$ is a Fibonacci number. Therefore, $A(10)=144$.
Thus, the probability $P=\frac{144}{1024}=\frac{9}{64}$, so $i+1=73$.