2021 Fall AMC 10B Problem 16

After Chris chooses an adjacent pair, Silva has $5$ equally likely adjacent pairs to choose. If Silva chooses the same pair, all $5$ balls return to their original positions. This has probability $\frac15.$ If Silva chooses a pair sharing exactly one ball with Chris’s pair, then $2$ balls are in their original positions. There are $2$ such pairs, so this has probability $\frac25.$ If Silva chooses a disjoint adjacent pair, then $1$ ball is in its original position. This also has probability $\frac25.$ The expected number is $$\begin{aligned} 5\cdot\frac15+2\cdot\frac25+1\cdot\frac25&=\frac{11}{5}\\ &=2.2. \end{aligned}$$ Thus, the answer is D .