Question

Juan 掷一枚公平的正八面体骰子，骰面标有 $1$ 到 $8$ 。然后 Amal 掷一枚公平的六面骰子。两次点数乘积是 $3$ 的倍数的概率是多少？

Juan rolls a fair regular octahedral die marked with the numbers $1$ through $8.$ Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of $3?$

$\dfrac{1}{12}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{7}{12}$

$\dfrac{2}{3}$

答案：C

知识点：对立事件概率骰子（概率）

难度评级：1430

解答：

乘积是 $3$ 的倍数，当且仅当至少一枚骰子掷出 $3$ 或 $6$ 。八面骰避开 $3,6$ 的概率为 $\dfrac68=\dfrac34$ ，六面骰避开它们的概率为 $\dfrac46=\dfrac23$ 。

因此两枚骰子都没有出现 $3$ 的倍数的概率是 $\dfrac34\cdot\dfrac23=\dfrac12$ ，所求概率为 $1-\dfrac12=\dfrac12$ 。

所以正确答案是 C。

The product is a multiple of $3$ if and only if at least one die shows $3$ or $6.$ The octahedral die avoids $3,6$ with probability $\dfrac68=\dfrac34,$ and the six-sided die avoids them with probability $\dfrac46=\dfrac23.$

So neither shows a multiple of $3$ with probability $\dfrac34\cdot\dfrac23=\dfrac12,$ and the answer is $1-\dfrac12=\dfrac12.$

Thus, the correct answer is C.

2002 AMC 12B 第 16 题

答案：C

难度评级：1430

解答：

其他年份的第 16 题