AMC12 2022 B

AMC12 2022 B · Q12

AMC12 2022 B · Q12. It mainly tests Inclusion–exclusion (basic), Probability (basic).

Kayla rolls four fair $6$-sided dice. What is the probability that at least one of the numbers Kayla rolls is greater than $4$ and at least two of the numbers she rolls are greater than $2$?

凯拉掷四个公平的 $6$ 面骰子。掷出至少一个数字大于 $4$ 且至少两个数字大于 $2$ 的概率是多少？

(A) \ \frac{2}{3} \ \frac{2}{3}

(B) \ \frac{19}{27} \ \frac{19}{27}

(D) \ \frac{61}{81} \ \frac{61}{81}

(E) \ \frac{7}{9} \ \frac{7}{9}

Answer

Correct choice: (D)

正确答案：（D）

Solution

We will subtract from one the probability that the first condition is violated and the probability that only the second condition is violated, being careful not to double-count the probability that both conditions are violated. For the first condition to be violated, all four dice must read $4$ or less, which happens with probability $\left( \frac23 \right)^4 = \frac{16}{81}$. For the first condition to be met but the second condition to be violated, at least one of the dice must read greater than $4$, but less than two of the dice can read greater than $2$. Therefore, one of the four die must read $5$ or $6$, while the remaining three dice must read $2$ or less, which happens with probability ${4 \choose 1} \left(\frac13\right) \left(\frac13\right)^3 = 4 \cdot \frac13 \cdot \frac{1}{27} = \frac{4}{81}$. Therefore, the overall probability of meeting both conditions is $1 - \frac{16}{81} - \frac{4}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}$.

我们将从 $1$ 中减去第一个条件被违反的概率和仅第二个条件被违反的概率，小心不要双重计算两个条件都被违反的概率。第一个条件被违反，即四个骰子都显示 $4$ 或更小，概率为 $\left( \frac23 \right)^4 = \frac{16}{81}$。第一个条件满足但第二个条件被违反，即至少一个骰子显示大于 $4$，但小于两个骰子显示大于 $2$。因此，四个骰子中有一个显示 $5$ 或 $6$，其余三个显示 $2$ 或更小，概率为 ${4 \choose 1} \left(\frac13\right) \left(\frac13\right)^3 = 4 \cdot \frac13 \cdot \frac{1}{27} = \frac{4}{81}$。因此，满足两个条件的总体概率为 $1 - \frac{16}{81} - \frac{4}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}$。

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