Difference between revisions of "2004 AMC 8 Problems/Problem 21"
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We can make a chart and the we see that the 12 possibilities: 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, and 12. Out of these only 8 work; thus the probability is <cmath> \boxed{\textbf{(D)}\ \frac23}</cmath> | We can make a chart and the we see that the 12 possibilities: 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, and 12. Out of these only 8 work; thus the probability is <cmath> \boxed{\textbf{(D)}\ \frac23}</cmath> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/yejMPaQ2uyY Soo, DRMS, NM | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=20|num-a=22}} | {{AMC8 box|year=2004|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:05, 25 March 2022
Problem
Spinners 
and 
are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
![[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label("$A$",(-1,1)); label("$B$",(2,1)); label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5)); [/asy]](http://latex.artofproblemsolving.com/2/b/8/2b8c0e9de9eb3c68e3938fd3467803e6bbb24ef3.png)
Solution
An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from 
. Multiply the independent probabilities of each spinner getting an odd number together and subtract it from .
![\[1-\frac24 \cdot \frac23 = 1- \frac13 = \boxed{\textbf{(D)}\ \frac23}\]](http://latex.artofproblemsolving.com/b/d/d/bddedf7295c213c9be69119bbcc7eb0e4d721daa.png)
Solution 2
We can make a chart and the we see that the 12 possibilities: 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, and 12. Out of these only 8 work; thus the probability is ![\[\boxed{\textbf{(D)}\ \frac23}\]](http://latex.artofproblemsolving.com/c/b/b/cbb27c511fa00a1566a341e0d8bd7569b33ceae0.png)
Video Solution
https://youtu.be/yejMPaQ2uyY Soo, DRMS, NM
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
