Difference between revisions of "2009 AMC 8 Problems/Problem 12"
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==Problem== | ==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? | + | 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> | <asy> | ||
| Line 10: | Line 10: | ||
label("$1$",(0,.5)); | label("$1$",(0,.5)); | ||
label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); | label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); | ||
| − | label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2)); | + | label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));</asy> |
| − | + | <asy> | |
| + | unitsize(30); | ||
draw(unitcircle); | draw(unitcircle); | ||
draw((0,0)--(0,-1)); | draw((0,0)--(0,-1)); | ||
| Line 20: | Line 21: | ||
label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));</asy> | label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));</asy> | ||
| − | <math> \textbf{(A)}\ \frac {1}{2} \qquad \textbf{(B)}\ \frac {2}{3} \qquad \textbf{(C)}\ \frac {3}{4} \qquad \textbf{(D)}\ \frac {7}{9} \qquad \textbf{(E)}\ \frac {5}{6}</math> | + | <math> \textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{2}{3}\qquad\textbf{(C)}\ \frac{3}{4}\qquad\textbf{(D)}\ \frac{7}{9}\qquad\textbf{(E)}\ \frac{5}{6} </math> |
| + | |||
| + | |||
| + | ==Solution== | ||
| + | The possible sums are | ||
| + | <cmath>\begin{tabular}{c|ccc} | ||
| + | & 1 & 3 & 5 \\ \hline | ||
| + | 2 & 3 & 5 & 7 \\ | ||
| + | 4 & 5 & 7 & 9 \\ | ||
| + | 6 & 7 & 9 & 11 | ||
| + | \end{tabular}</cmath> | ||
| + | |||
| + | 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 ~David | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2009|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:36, 15 April 2023
Contents
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)
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 ~David
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. 
