1994 AJHSME Problems/Problem 22

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]

$\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}$

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 $\frac14$ and the chance of getting an odd is $\frac34$. On the second wheel, the chance of getting an even is $\frac23$ and an odd is $\frac13$. D is correct, so take note.

\[\frac14 \cdot \frac23 + \frac34 \cdot \frac13 = \frac16 + \frac14 = \boxed{\text{(D)}\ \frac{5}{12}}\]

See Also

1994 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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