Difference between revisions of "2011 AMC 10B Problems/Problem 13"
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Two real numbers are selected independently at random from the interval <math>[-20, 10]</math>. What is the probability that the product of those numbers is greater than zero? | Two real numbers are selected independently 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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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 <math>\frac{1}{3}</math> of the total interval, and the interval for a negative number is <math>\frac{2}{3}</math>. Therefore, the probability the product is greater than zero is | 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 <math>\frac{1}{3}</math> of the total interval, and the interval for a negative number is <math>\frac{2}{3}</math>. Therefore, the probability the product is greater than zero is | ||
<cmath>\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}}</cmath> | <cmath>\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}}</cmath> | ||
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| + | == See Also== | ||
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| + | {{AMC10 box|year=2011|ab=B|num-b=12|num-a=14}} | ||
Revision as of 16:14, 4 June 2011
Problem
Two real numbers are selected independently at random from the interval ![$[-20, 10]$](http://latex.artofproblemsolving.com/0/a/8/0a843f746fa1baae8bb70d171f959327f2651c2c.png)
. What is the probability that the product of those numbers is greater than zero?
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 
of the total interval, and the interval for a negative number is 
. Therefore, the probability the product is greater than zero is
![\[\frac{1}{3} \times \frac{1}{3} + \frac{2}{3} \times \frac{2}{3} = \frac{1}{9} + \frac{4}{9} = \boxed{\textbf{(D)} \frac{5}{9}}\]](http://latex.artofproblemsolving.com/a/7/4/a74a71fdd8e6514e00ac335ccd72565ca5c4f101.png)
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 10 Problems and Solutions | ||
