Question

从 $60$ 的正因数中随机抽取一个，它小于 $7$ 的概率是多少？

What is the probability that a randomly drawn positive factor of $60$ is less than $7?$

$\dfrac{1}{10}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$60$ 的因数为 $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30$ 和 $60$ 。

其中十二个因数有六个小于 $7$ ，所以概率为 $\dfrac{6}{12} = \dfrac{1}{2}$ 。

所以正确答案是 E。

The factors of $60$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30,$ and $60.$

Six of these twelve factors are less than $7,$ so the probability is $\dfrac{6}{12} = \dfrac{1}{2}.$

Thus, the correct answer is E.