jmc

Let $a_n$ denote the number of sequences of length $n$ that do not contain consecutive $1$s. A sequence of length $n$ must either end in a $0$ or a $1$. If the string of length $n$ ends in a $0$, this string could have been formed by appending a $0$ to any sequence of length $n-1$, of which there are $a_{n-1}$ such strings. If the string of length $n$ ends in a $1$, this string could have been formed by appending a $01$ (to avoid consecutive $1$s) to any sequence of length $n-2$, of which there are $a_{n-2}$ such strings. Thus, we have the recursion$$a_n=a_{n-1}+a_{n-2}$$Solving for initial conditions, we find $a_1=2,a_2=3$. Thus we have the Fibonacci sequence with shifted indices; indeed $a_n=F_{n+2}$, so $a_{10}=F_{12}=144$. The probability is $\frac{144}{2^{10}}=\frac{9}{64}$, and $m+n=\boxed{73}$.