2023 AMC 10B Problem 19

Attempt Problem 19 of the 2023 AMC 10B 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 2023 AMC 10B solutions, or check the answer key.

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19.

Sonya the frog chooses a point uniformly at random lying within the square [0,6]×[0,6][0, 6] \times [0, 6] in the coordinate plane and hops to that point. She then chooses a distance uniformly at random from [0,1][0, 1] and a direction uniformly at random from {north,south,east,west}.\{\text{north}, \text{south}, \text{east}, \text{west}\}. All her choices are independent. She now hops the distance in the chosen direction. What is the probability that she lands outside the square?

16\dfrac{1}{6}

112\dfrac{1}{12}

14\dfrac{1}{4}

110\dfrac{1}{10}

19\dfrac{1}{9}

Answer: B
Concepts:geometric probabilitysymmetry
Difficulty rating: 1990
Solution:

The four directions behave the same by symmetry, so say she hops east. She lands outside exactly when her xx-coordinate plus the hop distance dd tops 6.6. Fix d.d. Her xx-coordinate is uniform on [0,6],[0, 6], so it beats 6d6 - d with probability d6.\frac{d}{6}. Now average over dd uniform on [0,1]:[0, 1]: 1612=112.\frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}. Thus, B is the correct answer.

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