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Difference between revisions of "2011 AMC 12B Problems/Problem 9"

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<math>\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}</math>
 
<math>\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}</math>
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[[2011 AMC 12B Problems/Problem 9|Solution]]
 
  
 
==Solution==
 
==Solution==
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Both numbers are negative with a <math>\frac{2}{3}*\frac{2}{3}=\frac{4}{9}</math> chance.
 
Both numbers are negative with a <math>\frac{2}{3}*\frac{2}{3}=\frac{4}{9}</math> chance.
  
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Both numbers are positive with a <math>\frac{1}{3}*\frac{1}{2}=\frac{1}{9}</math> chance.
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Both numbers are positive with a <math>\frac{1}{3}*\frac{1}{3}=\frac{1}{9}</math> chance.
  
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Therefore, the total probability is <math>\frac{4}{9}+\frac{1}{9}=\frac{5}{9}</math> and we are done. <math>\textbf{(D)}</math>
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Therefore, the total probability is <math>\frac{4}{9}+\frac{1}{9}=\frac{5}{9}</math> and we are done. <math>\boxed{D}</math>
  
  
 
{{AMC12 box|year=2011|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2011|ab=B|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 10:03, 4 July 2013

Problem

Two real numbers are selected independently and 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 to be greater than zero, we must have either both numbers negative or both positive.

Both numbers are negative with a $\frac{2}{3}*\frac{2}{3}=\frac{4}{9}$ chance.

Both numbers are positive with a $\frac{1}{3}*\frac{1}{3}=\frac{1}{9}$ chance.

Therefore, the total probability is $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$ and we are done. $\boxed{D}$


2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AMC 12 Problems and Solutions

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