Question

一个三位整数包含数字 $1$ 、 $3$ 、 $5$ 各一次。这个整数能被 $5$ 整除的概率是多少？

A three-digit integer contains one of each of the digits $1,$ $3,$ and $5.$ What is the probability that the integer is divisible by $5?$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{5}{6}$

一个数能被 $5$ 整除，当且仅当末位数字是 $0$ 或 $5$

A number is divisible by $5$ exactly when its last digit is $0$ or $5.$

对数字 $1$ 、 $3$ 、 $5$ ，每个数字作为末位的可能性相同

With digits $1,$ $3,$ and $5,$ each digit is equally likely to be last.

这个数以 $1$ 、 $3$ 或 $5$ 结尾的可能性相同。

只有当末位是 $5$ 时，它才能被 $5$ 整除，概率为 $\dfrac{1}{3}$ 。

所以正确答案是 B。

The number is equally likely to end in $1,$ $3,$ or $5.$

It is divisible by $5$ only if the last digit is $5,$ which happens with probability $\dfrac{1}{3}.$

Thus, B is the correct answer.