2004 AMC 12A Problem 20

Attempt Problem 20 of the 2004 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 2004 AMC 12A solutions, or check the answer key.

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

20.

Select numbers aa and bb between 00 and 11 independently and at random, and let cc be their sum. Let A,A, B,B, and CC be the results when a,a, b,b, and c,c, respectively, are rounded to the nearest integer. What is the probability that A+B=C?A + B = C?

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

Answer: E
Concepts:geometric probabilitycasework
Difficulty rating: 1990
Solution:

Represent the choices as a point (a,b)(a, b) in the unit square. Each of aa and bb rounds to 00 if below 12\tfrac12 and to 11 otherwise, while c=a+bc = a + b rounds based on 12\tfrac12 and 32.\tfrac32.

The equation fails in exactly two regions. If a,b<12,a, b \lt \tfrac12, it fails when a+b12;a + b \ge \tfrac12; this is a right triangle of area 18.\tfrac18. If a,b12,a, b \ge \tfrac12, it fails when a+b<32;a + b \lt \tfrac32; this is another right triangle of area 18.\tfrac18. When exactly one of a,ba, b is at least 12,\tfrac12, the equation always holds.

Thus the failure probability is 18+18=14,\tfrac18 + \tfrac18 = \tfrac14, so the requested probability is 114=34.1 - \tfrac14 = \tfrac34.

Thus, the correct answer is E.

← Problem 19#19
Full Exam

Problem 20 in Other Years