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2011 AMC 10B Problem 13

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2011 AMC 10B Problem 13AMC 10 Step by Step
Name:                             Date: September 13, 2026
Geometric Probability · difficulty 2 of 5 · about 2 min · position P11-15

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

  1. A)

    19\frac{1}{9}

  2. B)

    13\frac{1}{3}

  3. C)

    49\frac{4}{9}

  4. D)

    59\frac{5}{9}

  5. E)

    23\frac{2}{3}

Answer

D

Key insight

A positive product needs both numbers negative or both positive; the chances of negative and positive are 2/3 and 1/3, so add (2/3)^2 and (1/3)^2.

Solution

The interval [−20,10][-20, 10] has length 3030. A uniformly chosen number is negative with probability 2030=23\frac{20}{30} = \frac{2}{3} and positive with probability 1030=13\frac{10}{30} = \frac{1}{3} (being exactly 00 has probability 00).

The product is positive exactly when the two numbers have the same sign. Since the choices are independent,
P(both negative)+P(both positive)=(23)2+(13)2=49+19=59.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 (D) 59\boxed{\textbf{(D)}\ \frac{5}{9}}.

Why this works

Only the sign of each number matters, so the continuous problem collapses to a two-outcome experiment with probabilities given by lengths. The same-sign event splits into two disjoint pieces, each a product of independent probabilities.

Alternative approach

Complement: the product is negative when signs differ, with probability 2⋅23⋅13=492 \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{4}{9}, so the answer is 1−49=591 - \frac{4}{9} = \frac{5}{9}. In the 30×3030 \times 30 square of outcomes, the two same-sign regions have areas 400400 and 100100 out of 900900.

The trap

Assuming positive and negative are equally likely (answer 1/2) instead of using the interval lengths 20 and 10.

Common mistakes

  • Assuming positive and negative are equally likely (answer 1/21/2) instead of using the interval lengths 2020 and 1010.
  • Counting only the both-negative case, giving 4/94/9.

Techniques

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