Question

Abby、Bridget 和另外四位同学将坐成两排、每排三人来拍集体照，如图所示。如果座位随机分配，那么 Abby 和 Bridget 在同一行或同一列相邻的概率是多少？ $\begin{array}{ccc} \text{\text{X}}&\text{X}&\text{X} \\ \text{X}&\text{X}&\text{X} \end{array}$

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?

$\dfrac{1}{3}$

$\dfrac{2}{5}$

$\dfrac{7}{15}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

答案：C

知识点：基本概率分类讨论

难度评级：1210

视频讲解：

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文字解答：

把情况分成两类：第一类是 Abby 坐在中间列的两个座位之一，第二类是她坐在外侧的四个座位之一。

第一类发生的概率是 $\dfrac26 = \dfrac13$ 。此时 Bridget 要与 Abby 相邻，可以坐在 Abby 左右两个座位中的一个，或者坐在同一列的座位，共有 $3$ 个可选位置；剩下共有 $5$ 个空位，所以条件概率是 $\frac35$ 。因此这一类的概率为 $\dfrac13 \cdot \dfrac35 = \dfrac{3}{15} .$

第二类发生的概率是 $\dfrac46 = \dfrac23$ 。此时 Bridget 要与 Abby 相邻，只能坐在同一行相邻的那个座位，或同一列的座位，共有 $2$ 个可选位置；剩下仍有 $5$ 个空位，所以条件概率是 $\frac25$ 。因此这一类的概率为 $\frac23 \cdot \frac25 = \frac{4}{15} .$

因此任一类发生的总概率为 $\dfrac{3}{15} + \dfrac{4}{15} = \dfrac{7}{15}$ 。

所以正确答案是 C。

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 第 11 题

答案：C

难度评级：1210

视频讲解：

文字解答：

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