Difference between revisions of "2009 AMC 10B Problems/Problem 25"
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== Solution == | == Solution == | ||
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| + | === Solution 1 === | ||
| + | There are two possible stripe orientations for each of the six faces of the cube, so there are <math>2^6 = 64</math> possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of <math>3 \cdot 2 \cdot 2 = 12</math> stripe combinations on the cube result in a continuous stripe around the cube. The required probability is <math>\frac {12}{64} = \boxed{\frac {3}{16}}</math>. | ||
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| + | === Solution 2 === | ||
| + | Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces are painted to form an encircling stripe. The probability of the first case is <math>(\frac 12)^3 = \frac 18</math>, and the probability of the second case is <math>(\frac 12)^4 = \frac {1}{16}</math>. The cases are disjoint, so the probabilities sum <math>\frac 18 + \frac {1}{16} = \boxed {\frac {3}{16}}</math>. | ||
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| + | === Solution 3 === | ||
| + | There are three possible orientations of an encircling stripe. For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probability of each such stripe alignment is <math>(\frac 12)^4 = \frac {1}{16}</math>. Since there are three such possibilities and they are disjoint, the total probability is <math>3 \cdot \frac {1}{16} = \boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(B)}</math>. | ||
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| + | == See also == | ||
{{AMC10 box|year=2009|ab=B|num-b=24|after=Last question}} | {{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}} | ||
Revision as of 17:56, 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?
Solution
Solution 1
There are two possible stripe orientations for each of the six faces of the cube, so there are 
possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of 
stripe combinations on the cube result in a continuous stripe around the cube. The required probability is 
.
Solution 2
Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces are painted to form an encircling stripe. The probability of the first case is 
, and the probability of the second case is 
. The cases are disjoint, so the probabilities sum 
.
Solution 3
There are three possible orientations of an encircling stripe. For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probability of each such stripe alignment is . Since there are three such possibilities and they are disjoint, the total probability is 
. The answer is 
.
See also
| 2009 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 (Problems • Answer Key • Resources) | |
| 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 | |
