1987 AMC 12 Problem 26

Question

The amount

2.5

is split into two nonnegative real numbers uniformly at random, for instance, into

2.143

and

0.357,

or into

\sqrt3

and

2.5-\sqrt3.

Then each number is rounded to its nearest integer, for instance,

2

and

0

in the first case above,

2

and

1

in the second. What is the probability that the two integers sum to

3?

\frac14

\frac25

\frac12

\frac35

\frac34

Let the first number be

x

uniformly distributed on

[0,2.5]

Find the intervals on which

x

and

2.5-x

round to integers whose sum is

3

Let the first part be

x\in[0,2.5],

so the second is

2.5-x.

Ignoring endpoints of probability zero, the rounded values sum to

3

exactly when

\begin{gathered} \frac12\lt x\lt1\\ \text{or}\\ \frac32\lt x\lt2. \end{gathered}

These intervals have total length

1,

out of a sample interval of length

2.5.

The probability is therefore

\frac{1}{2.5}=\frac{2}{5}.

Thus the correct answer is B.

Problem 26 in other years