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

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==Problem==
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Two real numbers are selected independently and at random from the interval <math>[-20,10]</math>.  What is the probability that the product of those numbers is greater than zero?
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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>
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[[2011 AMC 12B Problems/Problem 9|Solution]]
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==Solution==
 
For the product to be greater than zero, we must have either both numbers negative or both positive.  
 
For the product to be greater than zero, we must have either both numbers negative or both positive.  
  
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Both numbers are positive with a <math>\frac{1}{3}*\frac{1}{2}=\frac{1}{9}</math> chance.
 
Both numbers are positive with a <math>\frac{1}{3}*\frac{1}{2}=\frac{1}{9}</math> chance.
  
Therefore, the total probability is <math>\frac{4}{9}+\frac{1}{9}=\frac{5}{9}</math> and we are done. <math>\Box</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>\textbf{(D)}</math>
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{{AMC12 box|year=2011|ab=B|num-b=8|num-a=10}}

Revision as of 18:31, 29 May 2011

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

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 $\frac{1}{3}*\frac{1}{2}=\frac{1}{9}$ chance.

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


2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions