Question

掷一枚不公平骰子时，出现偶数的可能性是出现奇数的 $3$ 倍。掷这枚骰子两次，掷出点数和为偶数的概率是多少？

When a certain unfair die is rolled, an even number is $3$ times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?

$\dfrac{3}{8}$

$\dfrac{4}{9}$

$\dfrac{5}{9}$

$\dfrac{9}{16}$

$\dfrac{5}{8}$

答案：E

知识点：骰子（概率）奇偶性独立事件

难度评级：900

解答：

设掷出奇数的概率为 $p$ ，则掷出偶数的概率为 $3p$ 。 $p + 3p = 1 \Rightarrow p = \dfrac{1}{4}.$

和为偶数当且仅当两次结果奇偶性相同，其概率为 $\dfrac{1}{4}^2 + \dfrac{3}{4}^2 = \dfrac{10}{16} = \dfrac{5}{8}.$

所以正确答案是 E。

Let $p$ be the probability that an odd number is rolled. Then $3p$ is the probability an even number is rolled. We know that $p + 3p = 1 \Rightarrow p = \dfrac{1}{4}.$

The only way for the sum to be even is if both rolls have the same parity. This happens with a probability of $\dfrac{1}{4}^2 + \dfrac{3}{4}^2 = \dfrac{10}{16} = \dfrac{5}{8}.$

Thus, E is the correct answer.

2021 AMC 10A Fall 第 9 题

答案：E

难度评级：900

解答：

其他年份的第 9 题