先放最明确的重叠：$D\cap R=0$，$D\cap C=2$。再用各集合总数倒推出只在单个集合或其他重叠中的人数。

Since no family had a dog and a rabbit,

Also,

Since $n(D)=12$, the dog-only region is

There are $9$ families outside all three circles.

The total number inside at least one circle is

Using the other totals, the completed regions are:

The completed diagram is

互斥表示没有交集。题目直接给了 no family had a dog and a rabbit。

$D$ and $R$ are mutually exclusive, since

恰好两种宠物，对应 $D\cap C$ 和 $C\cap R$ 两块，因为 $D\cap R=0$。

The number of families with exactly 2 kinds of pets is

So the probability is

检验独立性可以比较 $P(D\cap C)$ 与 $P(D)P(C)$。

Also,

and

So

Since

the events $D$ and $C$ are not independent.

条件概率独立性判定法。 除了使用乘积公式 $P(D \cap C) = P(D) P(C)$ 检验独立性外，也可以利用条件概率进行判定：

本解法通过计算在有狗的前提下家庭有猫的条件概率 $P(C \mid D)$，并与无条件概率 $P(C)$ 作比较。如果两者相等，则说明 $D$ 发生与否不改变 $C$ 发生的概率，即两事件独立；若不等，则不独立。这种方法直接从独立性的实际统计学意义出发，计算形式也更为简单。

We can test for independence by checking if $P(C \mid D) = P(C)$ (or equivalently, $P(D \mid C) = P(D)$).

From the Venn diagram, we have:

Calculate the conditional probability $P(C \mid D)$:

Since:

the events $D$ and $C$ are not independent.

(Note: Alternatively, we can calculate $P(D \mid C) = \frac{P(C \cap D)}{P(C)} = \frac{2}{14} = \frac{1}{7}$. Since $P(D \mid C) \neq P(D) = \frac{3}{10}$, they are not independent.)

已知 Sarah 家有 cat，所以分母是所有有 cat 的 14 家；其中也有 rabbit 的是 4 家。

条件是没有 cat。没有 cat 的家庭共有 $10+7+9=26$ 家，其中有 dog 或 rabbit 的是 $10+7=17$ 家。

Among families without a cat, the total number is

Among these, the number with a dog or a rabbit is

Therefore