Question

Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. $\begin{array}{ccc} \text{\text{X}}&\text{X}&\text{X} \\ \text{X}&\text{X}&\text{X} \end{array}$ If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?

$\dfrac{1}{3}$

$\dfrac{2}{5}$

$\dfrac{7}{15}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Video solution:

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Written solution:

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 on the left or the two seats on the right of Abby, 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.

2018 AMC 8 Problem 11

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Written solution:

Problem 11 in Other Years