Difference between revisions of "2009 AMC 10B Problems/Problem 25"

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{{duplicate|[[2009 AMC 12B Problems|2009 AMC 12B #17]] and [[2009 AMC 10B Problems|2009 AMC 10B #25]]}}
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{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #25]] and [[2009 AMC 12B Problems|2009 AMC 12B #17]]}}
  
 
== Problem ==
 
== Problem ==
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<math>\boxed{\frac {3}{16}}</math>.  The answer is <math>\mathrm{(B)}</math>.
 
<math>\boxed{\frac {3}{16}}</math>.  The answer is <math>\mathrm{(B)}</math>.
  
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{{AMC10 box|year=2009|ab=B|num-b=24|after=Last question}}
 
{{AMC12 box|year=2009|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2009|ab=B|num-b=16|num-a=18}}
{{AMC10 box|year=2009|ab=B|num-b=24|after=Last question}}
 

Revision as of 14:43, 26 February 2009

The following problem is from both the 2009 AMC 10B #25 and 2009 AMC 12B #17, so both problems redirect to this page.

Problem

Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?

$\mathrm{(A)}\frac 18\qquad \mathrm{(B)}\frac {3}{16}\qquad \mathrm{(C)}\frac 14\qquad \mathrm{(D)}\frac 38\qquad \mathrm{(E)}\frac 12$

Solution

$\boxed{\frac {3}{16}}$. The answer is $\mathrm{(B)}$.

2009 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last question
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 AMC 10 Problems and Solutions
2009 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 AMC 12 Problems and Solutions