Difference between revisions of "1994 AJHSME Problems/Problem 22"
Mrdavid445 (talk | contribs) (Created page with "==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)...") |
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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>. | ||
| + | |||
| + | <cmath>\frac14 \cdot \frac23 + \frac34 \codt \frac13 = \frac16 + \frac14 = \boxed{\text{(D)}\ \frac{5}{12}}</cmath> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=21|num-a=23}} | ||
Revision as of 01:09, 23 December 2012
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 
.
\[\frac14 \cdot \frac23 + \frac34 \codt \frac13 = \frac16 + \frac14 = \boxed{\text{(D)}\ \frac{5}{12}}\] (Error compiling LaTeX. Unknown error_msg)
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 | ||
