Difference between revisions of "2006 AMC 8 Problems/Problem 17"
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Revision as of 00:22, 5 July 2013
Problem
Jeff rotates spinners 
, 
and 
and adds the resulting numbers. What is the probability that his sum is an odd number?
![[asy] size(200); path circle=circle((0,0),2); path r=(0,0)--(0,2); draw(circle,linewidth(1)); draw(shift(5,0)*circle,linewidth(1)); draw(shift(10,0)*circle,linewidth(1)); draw(r,linewidth(1)); draw(rotate(120)*r,linewidth(1)); draw(rotate(240)*r,linewidth(1)); draw(shift(5,0)*r,linewidth(1)); draw(shift(5,0)*rotate(90)*r,linewidth(1)); draw(shift(5,0)*rotate(180)*r,linewidth(1)); draw(shift(5,0)*rotate(270)*r,linewidth(1)); draw(shift(10,0)*r,linewidth(1)); draw(shift(10,0)*rotate(60)*r,linewidth(1)); draw(shift(10,0)*rotate(120)*r,linewidth(1)); draw(shift(10,0)*rotate(180)*r,linewidth(1)); draw(shift(10,0)*rotate(240)*r,linewidth(1)); draw(shift(10,0)*rotate(300)*r,linewidth(1)); label("$P$", (-2,2)); label("$Q$", shift(5,0)*(-2,2)); label("$R$", shift(10,0)*(-2,2)); label("$1$", (-1,sqrt(2)/2)); label("$2$", (1,sqrt(2)/2)); label("$3$", (0,-1)); label("$2$", shift(5,0)*(-sqrt(2)/2,sqrt(2)/2)); label("$4$", shift(5,0)*(sqrt(2)/2,sqrt(2)/2)); label("$6$", shift(5,0)*(sqrt(2)/2,-sqrt(2)/2)); label("$8$", shift(5,0)*(-sqrt(2)/2,-sqrt(2)/2)); label("$1$", shift(10,0)*(-0.5,1)); label("$3$", shift(10,0)*(0.5,1)); label("$5$", shift(10,0)*(1,0)); label("$7$", shift(10,0)*(0.5,-1)); label("$9$", shift(10,0)*(-0.5,-1)); label("$11$", shift(10,0)*(-1,0)); [/asy]](http://latex.artofproblemsolving.com/2/4/0/2402672f0bbb2b057f89b1a2623c46ea6f2d380c.png)
Solution
In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even + Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner contains only even numbers. Therefore we must work with Even + Even + Odd and spinner will give us one of our even numbers. We also see that spinner only contains odd, so spinner must give us our odd nmber. We still need one even number from spinner . There is only 1 even number: 
. Since spinning the required numbers are automatic on the other spinners, we only have to find the probability of spinning a in spinner , which clearly is 
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
