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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==Solution== | ==Solution== | ||
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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>\ | + | 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}} | ||
Revision as of 08:43, 12 July 2011
Problem
Two real numbers are selected independently and at random from the interval ![$[-20,10]$](http://latex.artofproblemsolving.com/d/f/e/dfeb28af51c3420c077378ff48d0d6e66a00ecd0.png)
. What is the probability that the product of those numbers is greater than zero?
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 
chance.
Both numbers are positive with a 
chance.
Therefore, the total probability is 
and we are done. 
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
