2023 AMC 10A Problem 7

The total can only reach exactly $3$ through the opening rolls, and these ways are disjoint: $3$ alone (probability $\frac16$), then $1,2$ and $2,1$ (each $\frac1{36}$), and $1,1,1$ (probability $\frac1{216}$). Add them up: $\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}.$ Thus, B is the correct answer.