Question

Select numbers $a$ and $b$ between $0$ and $1$ independently and at random, and let $c$ be their sum. Let $A,$ $B,$ and $C$ be the results when $a,$ $b,$ and $c,$ respectively, are rounded to the nearest integer. What is the probability that $A + B = C?$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

Answer: E

Concepts:geometric probability casework

Difficulty rating: 1990

Small Hint:

Model $(a, b)$ as a random point in the unit square and split by whether each of $a, b$ is below $\tfrac12$

Big Hint:

The condition fails only when $a, b \ge \tfrac12$ but $a + b \lt \tfrac32,$ or $a, b \lt \tfrac12$ but $a + b \ge \tfrac12$

Solution:

Represent the choices as a point $(a, b)$ in the unit square. Each of $a$ and $b$ rounds to $0$ if below $\tfrac12$ and to $1$ otherwise, while $c = a + b$ rounds based on $\tfrac12$ and $\tfrac32.$

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

Thus the failure probability is $\tfrac18 + \tfrac18 = \tfrac14,$ so the requested probability is $1 - \tfrac14 = \tfrac34.$

Thus, the correct answer is E.

2004 AMC 12A Problem 20

Answer: E

Difficulty rating: 1990

Small Hint:

Big Hint:

Solution:

Problem 20 in Other Years