Difference between revisions of "2009 AMC 8 Problems/Problem 12"
(→Problem) |
(→Solution) |
||
| Line 37: | Line 37: | ||
Only <math>9</math> is not prime, so there are <math>7</math> prime numbers and <math>9</math> total numbers for a probability of <math>\boxed{\textbf{(D)}\ \frac79}</math>. | Only <math>9</math> is not prime, so there are <math>7</math> prime numbers and <math>9</math> total numbers for a probability of <math>\boxed{\textbf{(D)}\ \frac79}</math>. | ||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=NPTaWKEkaHs | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=11|num-a=13}} | {{AMC8 box|year=2009|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:13, 17 June 2022
Problem
The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
![[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$1$",(0,.5)); label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]](http://latex.artofproblemsolving.com/e/f/2/ef23ec00d65a4669943af5033ead69ca82d03505.png)
![[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$2$",(0,.5)); label("$4$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]](http://latex.artofproblemsolving.com/3/0/5/3058f62f5eba87c82939ed3a26d69037c009ba5c.png)
Video Solution
https://www.youtube.com/watch?v=NPTaWKEkaHs
Solution
The possible sums are
![\[\begin{tabular}{c|ccc} & 1 & 3 & 5 \\ \hline 2 & 3 & 5 & 7 \\ 4 & 5 & 7 & 9 \\ 6 & 7 & 9 & 11 \end{tabular}\]](http://latex.artofproblemsolving.com/9/7/2/9726458cb656f3923af8d189b0010f6a38c19dd0.png)
Only 
is not prime, so there are 
prime numbers and total numbers for a probability of 
.
Video Solution
https://www.youtube.com/watch?v=NPTaWKEkaHs
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
