2011 AMC 10B Problem 13

P11-15~2 minPrint

Geometric Probability Basic Probability Set up the equation/formula and compute; no special trick needed

Two real numbers are selected independently at random from the interval $[-20, 10]$ . What is the probability that the product of those numbers is greater than zero?

Show solution without answering

Key insight

The product is positive when both numbers are negative or both are positive. A number is negative 2/3 of the time and positive 1/3 of the time.

Solution

Only the signs of the two numbers matter for whether the product is positive, so first find how likely each sign is.

The interval from $-20$ to $10$ is $30$ long.
The negative part, from $-20$ to $0$ , is $20$ long. So a number is negative with probability $\frac{20}{30} = \frac{2}{3}$ .
The positive part, from $0$ to $10$ , is $10$ long. So a number is positive with probability $\frac{10}{30} = \frac{1}{3}$ . (Landing exactly on $0$ has probability $0$ , so we can ignore it.)
A product is positive when the two numbers have the same sign.
Both negative: $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$ .
Both positive: $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$ .
These two cases cannot happen at once, so add them: $\frac{4}{9} + \frac{1}{9} = \frac{5}{9}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{5}{9}}$ .

Why this works

For a number picked evenly from an interval, the chance it lands in a piece is the piece's length divided by the whole length. The two picks do not affect each other, so you multiply their chances. Asking "which signs make the product positive?" turns a question about infinitely many numbers into a simple two-outcome question.

Alternative approach

Find the chance the product is negative and subtract from $1$ . A negative product needs one negative and one positive number, in either order: $\frac{2}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{3} = \frac{4}{9}$ . So the answer is $1 - \frac{4}{9} = \frac{5}{9}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The interval $[-20, 10]$ has length $30$ . A uniformly chosen number is negative with probability $\frac{20}{30} = \frac{2}{3}$ and positive with probability $\frac{10}{30} = \frac{1}{3}$ (being exactly $0$ has probability $0$ ).

The product is positive exactly when the two numbers have the same sign. Since the choices are independent,
$P(\text{both negative}) + P(\text{both positive}) = \left(\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2 = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}.$

The answer is $\boxed{\textbf{(D)}\ \frac{5}{9}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 10: Geometric Probability
Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques