2017 AMC 12A Problem 10

Attempt Problem 10 of the 2017 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

Chloé chooses a real number uniformly at random from the interval [0,2017].[0,2017]. Independently, Laurent chooses a real number uniformly at random from the interval [0,4034].[0,4034]. What is the probability that Laurent’s number is greater than Chloé’s number?
12\dfrac{1}{2}
23\dfrac{2}{3}
34\dfrac{3}{4}
56\dfrac{5}{6}
78\dfrac{7}{8}
Answer: C
Concepts:geometric probabilitysymmetrycomplementary probability
Small hint:
Split on whether Laurent’s number is above or below 20172017
Big hint:
If it is above 20172017 he always wins; if below, it is a symmetric coin flip
Solution:
With probability 12,\dfrac{1}{2}, Laurent’s number lies in [2017,4034],[2017,4034], which exceeds any number Chloé could choose, so he wins for certain.
With the other probability 12,\dfrac{1}{2}, Laurent’s number lies in [0,2017],[0,2017], matching Chloé’s interval; by symmetry he is larger half the time. The total probability is 12⋅1+12⋅12=34. \dfrac{1}{2}\cdot1+\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}.
Thus, the correct answer is C.
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