Question

一对公平的 $6$ 面骰子掷 $n$ 次。使得至少有一次掷出的点数和为 $7$ 的概率大于 $\dfrac{1}{2}$ 的最小 $n$ 是多少？

A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\dfrac{1}{2}?$

$2$

$3$

$4$

$5$

$6$

答案：C

知识点：骰子（概率）对立事件概率

难度评级：960

解答：

也可以求最小的 $n$ ，使没有掷出和为 $7$ 的概率小于 $\dfrac 12$ 每次以 $\dfrac 16$ 的概率掷出和为 $7$ 因此以 $\dfrac 56$ 的概率不掷出 $7$ 。

因此 $7$ 一次也没有出现的概率是 $\left(\dfrac 56\right)^n$ 要找最小的 $n$ ，使 $\left(\dfrac 56\right)^n < \dfrac 12$ 。

当 $n=3$ 时，概率为 $\dfrac{125}{216}$ ，大于 $\dfrac 12$ 。

当 $n=4$ 时，概率为 $\dfrac{625}{1296}$ ，小于 $\dfrac 12$ 因此答案是 $4$ 。

所以正确答案是 C。

To compute this, we can also find the least $n$ such that the probability of not rolling a $7$ is less than $\dfrac 12.$ Each roll has an independent probability of $\dfrac 16$ of getting $7,$ so it has a $\dfrac 56$ probability of not landing on $7.$

Thus, the probability of none of the rolls being $7$ is $\left(\dfrac 56\right)^n.$ We must find the least $n$ such that $\left(\dfrac 56\right)^n < \dfrac 12.$

If $n=3,$ then the probability is $\dfrac{125}{216},$ which is greater than $\dfrac 12.$

If $n=4,$ then the probability is $\dfrac{625}{1296},$ which is less than $\dfrac 12.$ This makes the answer $4.$

Thus, the answer is C .

2022 AMC 10B 第 12 题

答案：C

难度评级：960

解答：

其他年份的第 12 题