AMC10 2017 A

AMC10 2017 A · Q15

AMC10 2017 A · Q15. It mainly tests Probability (basic), Geometric probability (basic).

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. The other half of the time, he will pick a number between 0 and 2017, and by symmetry his number will be the larger one in half of those cases. Therefore the requested probability is\n\[\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}.\]

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。另一半时间，他选择 0 到 2017 之间的数，由对称性，他的数较大占一半情况。因此所需概率为 \[\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}\]。

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