Difference between revisions of "1989 AJHSME Problems/Problem 25"

Revision as of 05:57, 16 April 2012

Problem

Every time these two wheels are spun, two numbers are selected by the pointers. What is the probability that the sum of the two selected numbers is even?

$\text{(A)}\ \frac{1}{6} \qquad \text{(B)}\ \frac{3}{7} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{2}{3} \qquad \text{(E)}\ \frac{5}{7}$

$[asy] unitsize(36); draw(circle((-3,0),1)); draw(circle((0,0),1)); draw((0,0)--dir(30)); draw((0,0)--(0,-1)); draw((0,0)--dir(150)); draw((-2.293,.707)--(-3.707,-.707)); draw((-2.293,-.707)--(-3.707,.707)); fill((-2.9,1)--(-2.65,1.25)--(-2.65,1.6)--(-3.35,1.6)--(-3.35,1.25)--(-3.1,1)--cycle,black); fill((.1,1)--(.35,1.25)--(.35,1.6)--(-.35,1.6)--(-.35,1.25)--(-.1,1)--cycle,black); label("$5$",(-3,.2),N); label("$3$",(-3.2,0),W); label("$4$",(-3,-.2),S); label("$8$",(-2.8,0),E); label("$6$",(0,.2),N); label("$9$",(-.2,.1),SW); label("$7$",(.2,.1),SE); [/asy]$

Solution

For the sum to be even, the two selected numbers must have the same parity.

The first spinner has $2$ odd numbers and $2$ even, so no matter what the second spinner is, there is a $1/2$ chance the first spinner lands on a number with the same parity, so the probability of an even sum is $1/2\rightarrow \boxed{\text{C}}$ .

@@ Line 27: / Line 27: @@
 {{AJHSME box|year=1989|num-b=24|after=Last<br>Problem}}
 [[Category:Introductory Number Theory Problems]]
-[[Category:Probability Problems]]
+[[Category:Introductory Probability Problems]]

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Difference between revisions of "1989 AJHSME Problems/Problem 25"

Revision as of 05:57, 16 April 2012

Problem

Solution

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