2011 AMC 10B Problems/Problem 13

Revision as of 18:40, 29 October 2019 by Bluelinfish (talk | contribs) (Solution 2)

Problem

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?

$\textbf{(A)}\ \frac{1}{9} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{4}{9} \qquad\textbf{(D)}\ \frac{5}{9} \qquad\textbf{(E)}\ \frac{2}{3}$

Solution

For the product of two numbers to be greater than zero, they either have to both be negative or both be positive. The interval for a positive number is $\frac{1}{3}$ of the total interval, and the interval for a negative number is $\frac{2}{3}$. Therefore, the probability the product is greater than zero is \[\frac{1}{3} \times \frac{1}{3} + \frac{2}{3} \times \frac{2}{3} = \frac{1}{9} + \frac{4}{9} = \boxed{\textbf{(D)} \frac{5}{9}}\]


Solution 2

We can also plot this problem on the coordinate grid. First, plot the possible region [the square with endpoints $(10,10), (-20, 10), (-20, -20), (-20, 10)$] Then, we can find our possible region which is if both numbers are positive or if both numbers are negative. Lets take the first case in which both numbers are positive: We can plot this as the top right of our possible square [$(10,10) ,(0,0) ,(10,0) ,(0,10)$]. Similarly if both numbers are negative we get the square [$(-20,-20), (0,0), (-20,0) ,(0,-20)$] Then we find the area of the possible region $30\cdot 30 = 900$ and then find the areas of our positive regions $10\cdot 10=100$ and $20\cdot 20 = 400$. Adding and simplifying we get $\frac{100+400}{900}=\frac{5}{9}$.

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 10 Problems and Solutions

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