2015 AMC 10A Problem 22

Count the possible sets of people who stand. For $0$ and $1$ people standing, there are $1$ and $8$ possibilities. For $2$ people standing, choose any pair and subtract the $8$ adjacent pairs: $\binom82-8=20$. For $3$ people standing, first choose one standing person. Among the remaining five non-neighbor seats, $10$ pairs are possible, but $4$ of those pairs are adjacent, leaving $6$. This counts each final set three times, so there are $\frac{8\cdot6}{3}=16$ possibilities. For $4$ people standing, the only possibilities are the two alternating sets. Thus the number of favorable coin-flip outcomes is $1+8+20+16+2=47$. Since all $2^8=256$ outcomes are equally likely, the probability is $\frac{47}{256}$. Thus, A is the correct answer.