Question

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?

$\dfrac{1}{9}$

$\dfrac{1}{3}$

$\dfrac{4}{9}$

$\dfrac{5}{9}$

$\dfrac{2}{3}$

Solution:

There is $0$ probability that our number is $0,$ so we need to just find the probability that the product isn't less than $0.$ The number product is less than zero if one of the numbers is less than $0$ and one of them is greater than $0.$

First, there are $2$ ways to choose the designated lower number. Then, the probability that the designated lower number is less than $0$ is $\frac 23$ and the probability that the designated higher number is greater than $0$ is $\frac 13.$

This makes the probability that the product is less than $0$ equal to $2 \cdot \dfrac 23 \cdot \dfrac 13 = \dfrac 49.$ As such, the probability that the product is greater than $0$ equal to $\dfrac 59.$

Thus, the correct answer is D .

2011 AMC 10B Problem 13

Solution:

Problem 13 in Other Years