2004 AMC 12A Problem 20

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.