AMC12 2017 A

AMC12 2017 A · Q10

AMC12 2017 A · Q10. It mainly tests Probability (basic), Counting & probability misc.

Chloé chooses a real number uniformly at random from the interval $[0, 2017]$. Independently, Laurent chooses a real number uniformly at random from the interval $[0, 4034]$. What is the probability that Laurent’s number is greater than Chloé’s number?

Chloé 从区间 $[0, 2017]$ 中均匀随机选择一个实数。独立地，Laurent 从区间 $[0, 4034]$ 中均匀随机选择一个实数。Laurent 的数大于 Chloé 的数的概率是多少？

(A) \frac{1}{2} \frac{1}{2}

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

(D) \frac{5}{6} \frac{5}{6}

(E) \frac{7}{8} \frac{7}{8}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Half of the time Laurent will pick a number between 2017 and 4034, in which case the probability that his number will be greater than Chloé's number is 1. Half of the time Laurent will pick a number between 0 and 2017, in which case the probability that his number will be greater than Chloé's number is $\frac{1}{2}$ by symmetry. Therefore the requested probability is $\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}$.

Laurent 有半数时间选择 2017 到 4034 之间的数，此时他的数大于 Chloé 的数的概率为 1。另半数时间他选择 0 到 2017 之间的数，此时由对称性他的数大于 Chloé 的数的概率为 $\frac{1}{2}$。因此所求概率为 $\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}$。

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