2018 AMC 8 Problem 11

First choose Abby’s seat, then count how many seats are adjacent to it.

Middle seats and corner seats have different numbers of adjacent seats.

We can split the problem into two cases. In case $1,$ Abby is in one of the middle two seats, and in case $2,$ she is in one of the outer $4$ seats.

Firstly notice that there is a $\dfrac26 = \dfrac13$ probability of case $1$ being true (i.e. Abby is in the middle two seats). For Bridget to be adjacent to Abby in this case, she must be in either of the two seats beside Abby in the same row, or she is in the same column as her. There are $3$ ways to make this happen out of a possible $5$ open seats, so there is a $\frac35$ chance of this happening. Therefore, the total probability of this case is $$\dfrac13 \cdot \dfrac35 = \dfrac{3}{15} .$$

Next, notice that there is a $\dfrac46 = \dfrac23$ probability of case $2$ being true (i.e. Abby is in the outer four seats). For Bridget to be adjacent to Abby in this case, she must either be in the single seat next to Abby in the same row, or she is in the same column as Abby. There are $2$ ways to make this happen out of a possible $5$ open seats, so there is a $\frac25$ chance of this happening. Therefore, the total probability of this case is $$\frac23 \cdot \frac25 = \frac{4}{15} .$$

Therefore, the final probability of either of these cases happening is $\dfrac{3}{15} + \dfrac{4}{15} = \dfrac{7}{15} .$

Thus, C is the correct answer.