Difference between revisions of "1994 AJHSME Problems/Problem 22"
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<math>\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}</math> | <math>\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | An even sum occurs when an even is added to an even or an odd is added to an odd. Looking at the areas of the regions, the chance of getting an even in the first wheel is <math>\frac14</math> and the chance of getting an odd is <math>\frac34</math>. On the second wheel, the chance of getting an even is <math>\frac23</math> and an odd is <math>\frac13</math> , resulting in an answer of D. | ||
| + | |||
| + | <cmath>\frac14 \cdot \frac23 + \frac34 \cdot \frac13 = \frac16 + \frac14 = \boxed{\text{(D)}\ \frac{5}{12}}</cmath> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=21|num-a=23}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:38, 24 April 2021
Problem
The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is
![[asy] draw(circle((0,0),3)); draw(circle((7,0),3)); draw((0,0)--(3,0)); draw((0,-3)--(0,3)); draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2)); draw((7,0)--(7-3*sqrt(3)/2,-3/2)); draw((0,5)--(0,3.5)--(-0.5,4)); draw((0,3.5)--(0.5,4)); draw((7,5)--(7,3.5)--(6.5,4)); draw((7,3.5)--(7.5,4)); label("$3$",(-0.75,0),W); label("$1$",(0.75,0.75),NE); label("$2$",(0.75,-0.75),SE); label("$6$",(6,0.5),NNW); label("$5$",(7,-1),S); label("$4$",(8,0.5),NNE); [/asy]](http://latex.artofproblemsolving.com/8/3/2/8329d1cf2b3091dbd5a5cc6cf97299841af22092.png)
Solution
An even sum occurs when an even is added to an even or an odd is added to an odd. Looking at the areas of the regions, the chance of getting an even in the first wheel is 
and the chance of getting an odd is 
. On the second wheel, the chance of getting an even is 
and an odd is 
, resulting in an answer of D.
![\[\frac14 \cdot \frac23 + \frac34 \cdot \frac13 = \frac16 + \frac14 = \boxed{\text{(D)}\ \frac{5}{12}}\]](http://latex.artofproblemsolving.com/8/d/a/8da92819d08fca6fd8ed07eab95a4438f08f460d.png)
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
