AMC10 2022 B

AMC10 2022 B · Q12

AMC10 2022 B · Q12. It mainly tests Probability (basic).

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 $\frac{1}{2}$?

一对公平的$6$面骰子掷$n$次。求最小的$n$，使得至少一次掷骰和为$7$的概率大于$\frac{1}{2}$。

(A) 2 2

(B) 3 3

(D) 5 5

(E) 6 6

Answer

Correct choice: (C)

正确答案：（C）

Solution

Rolling a pair of fair $6$-sided dice, the probability of getting a sum of $7$ is $\frac16:$ Regardless what the first die shows, the second die has exactly one outcome to make the sum $7.$ We consider the complement: The probability of not getting a sum of $7$ is $1-\frac16=\frac56.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-\left(\frac56\right)^n.$ Therefore, we have $1-\left(\frac56\right)^n>\frac12,$ or \[\left(\frac56\right)^n<\frac12.\] Since $\left(\frac56\right)^4<\frac12<\left(\frac56\right)^3,$ the least integer $n$ satisfying the inequality is $\boxed{\textbf{(C) } 4}.$

掷一对公平的$6$面骰子，和为$7$的概率是$\frac{1}{6}$：无论第一个骰子显示什么，第二个骰子恰好有一个结果使和为$7$。考虑补事件：不得到和为$7$的概率是$1-\frac{1}{6}=\frac{5}{6}$。掷$n$次，和为$7$至少一次的概率是$1-\left(\frac{5}{6}\right)^n$。因此，$1-\left(\frac{5}{6}\right)^n>\frac{1}{2}$，即\[\left(\frac{5}{6}\right)^n<\frac{1}{2}\]由于$\left(\frac{5}{6}\right)^4<\frac{1}{2}<\left(\frac{5}{6}\right)^3$，满足不等式的最小整数$n$是\boxed{\textbf{(C) } 4}。

Topics

CP.006 Probability (basic)