2003 AMC 10A Problem 12

Below is the professionally curated solution for Problem 12 of the 2003 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10A solutions, or check the answer key.

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Concepts:geometric probability triangle area

Difficulty rating: 1410

12.

A point $(x, y)$ is randomly picked from inside the rectangle with vertices $(0, 0),$ $(4, 0),$ $(4, 1),$ and $(0, 1).$ What is the probability that $x \lt y?$

$\dfrac{1}{8}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

Solution:

The condition $x \lt y$ holds in the triangle bounded by $y = x,$ $y = 1,$ and $x = 0,$ which has vertices $(0, 0),$ $(0, 1),$ and $(1, 1).$

This triangle has area $\dfrac{1}{2},$ while the rectangle has area $4.$

The probability is $\dfrac{1/2}{4} = \dfrac{1}{8}.$

Thus, the correct answer is A.

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