Difference between revisions of "2011 AMC 10B Problems/Problem 13"
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== Solution 2 == | == Solution 2 == | ||
| − | We can also plot this problem on the coordinate grid. First, plot the possible region [the square with endpoints <math>(10,10), (-20, 10), (-20, -20), (-20, 10)</math>] 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 [<math>(10,10) ,(0,0) ,(10,0) ,(0,10)</math>]. Similarly if both numbers are negative we get the square [<math>(-20,-20), (0,0), (-20,0) ,(0,-20)</math>] Then we find the area of the possible region <math>30\cdot 30 = 900</math> and then find the areas of our positive regions <math>10\cdot 10=100 and 20\cdot 20 = 400. Adding and simplifying we get < | + | We can also plot this problem on the coordinate grid. First, plot the possible region [the square with endpoints <math>(10,10), (-20, 10), (-20, -20), (-20, 10)</math>] 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 [<math>(10,10) ,(0,0) ,(10,0) ,(0,10)</math>]. Similarly if both numbers are negative we get the square [<math>(-20,-20), (0,0), (-20,0) ,(0,-20)</math>] Then we find the area of the possible region <math>30\cdot 30 = 900</math> and then find the areas of our positive regions <math>10\cdot 10=100</math> and <math>20\cdot 20 = 400</math>. Adding and simplifying we get <math>\frac{100+400}{900}=\frac{5}{9}</math>. |
== See Also== | == See Also== | ||
Revision as of 17:40, 29 October 2019
Contents
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)
Solution 2
We can also plot this problem on the coordinate grid. First, plot the possible region [the square with endpoints 
] 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 [
]. Similarly if both numbers are negative we get the square [
] Then we find the area of the possible region 
and then find the areas of our positive regions 
and 
. Adding and simplifying we get 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
