AMC12 2005 A

AMC12 2005 A · Q14

AMC12 2005 A · Q14. It mainly tests Probability (basic), Conditional probability (basic).

On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

在一个标准骰子上，随机移除一个点，每个点被选中的概率相等。然后掷骰子。顶面点数为奇数的概率是多少？

(A) \frac{5}{11} \frac{5}{11}

(B) \frac{10}{21} \frac{10}{21}

(D) \frac{11}{21} \frac{11}{21}

(E) \frac{6}{11} \frac{6}{11}

Answer

Correct choice: (D)

正确答案：（D）

Solution

There are $1 + 2 + 3 + 4 + 5 + 6 = 21$ dots total. Casework: - The dot is removed from an even face. There is a $\frac{2+4+6}{21} = \frac{4}{7}$ chance of this happening. Then there are 4 odd faces, giving us a probability of $\frac 47 \cdot \frac 46 = \frac{8}{21}$. - The dot is removed from an odd face. There is a $\frac{1+3+5}{21} = \frac{3}{7}$ chance of this happening. Then there are 2 odd faces, giving us a probability of $\frac 37 \cdot \frac 26 = \frac{1}{7}$. Thus the answer is $\frac 8{21} + \frac 17 = \frac{11}{21} \Longrightarrow \mathrm{(D)}$.

总共有 $1 + 2 + 3 + 4 + 5 + 6 = 21$ 个点。分类讨论： - 移除的点来自偶数面。发生概率为 $\frac{2+4+6}{21} = \frac{4}{7}$。此时仍有 4 个奇数面，因此概率为 $\frac 47 \cdot \frac 46 = \frac{8}{21}$。 - 移除的点来自奇数面。发生概率为 $\frac{1+3+5}{21} = \frac{3}{7}$。此时只有 2 个奇数面，因此概率为 $\frac 37 \cdot \frac 26 = \frac{1}{7}$。因此答案为 $\frac 8{21} + \frac 17 = \frac{11}{21} \Longrightarrow \mathrm{(D)}$。

Topics