Difference between revisions of "2007 AMC 12B Problems/Problem 13"

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==Problem 13==
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== Problem ==
 
A traffic light runs repeatedly through the following cycle: green for <math>30</math> seconds, then yellow for <math>3</math> seconds, and then red for <math>30</math> seconds. Leah picks a random three-second time interval to watch the light. What is the probability that the color changes while she is watching?
 
A traffic light runs repeatedly through the following cycle: green for <math>30</math> seconds, then yellow for <math>3</math> seconds, and then red for <math>30</math> seconds. Leah picks a random three-second time interval to watch the light. What is the probability that the color changes while she is watching?
  
<math>\mathrm {(A)} \frac{1}{63}</math>  <math>\mathrm {(B)} \frac{1}{21}</math>  <math>\mathrm {(C)} \frac{1}{10}</math>  <math>\mathrm {(D)} \frac{1}{7}</math>  <math>\mathrm {(E)} \frac{1}{3}</math>
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<math>\mathrm{(A)}\ \frac{1}{63} \qquad \mathrm{(B)}\ \frac{1}{21} \qquad \mathrm{(C)}\ \frac{1}{10} \qquad \mathrm{(D)}\ \frac{1}{7} \qquad \mathrm{(E)}\ \frac{1}{3}</math>
 
 
==Solution==
 
  
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== Solution ==
 
The traffic light runs through a <math>63</math> second cycle.
 
The traffic light runs through a <math>63</math> second cycle.
  
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<math>\frac{9}{63} = \frac{1}{7} \Rightarrow \mathrm{(D)}</math>
 
<math>\frac{9}{63} = \frac{1}{7} \Rightarrow \mathrm{(D)}</math>
  
==See Also==
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== See Also ==
 
{{AMC12 box|year=2007|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2007|ab=B|num-b=12|num-a=14}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 00:03, 19 October 2020

Problem

A traffic light runs repeatedly through the following cycle: green for $30$ seconds, then yellow for $3$ seconds, and then red for $30$ seconds. Leah picks a random three-second time interval to watch the light. What is the probability that the color changes while she is watching?

$\mathrm{(A)}\ \frac{1}{63} \qquad \mathrm{(B)}\ \frac{1}{21} \qquad \mathrm{(C)}\ \frac{1}{10} \qquad \mathrm{(D)}\ \frac{1}{7} \qquad \mathrm{(E)}\ \frac{1}{3}$

Solution

The traffic light runs through a $63$ second cycle.

Letting $t=0$ reference the moment it turns green, the light changes at three different times: $t=30$, $t=33$, and $t=63$

This means that the light will change if the beginning of Leah's interval lies in $[27,30]$, $[30,33]$ or $[60,63]$

This gives a total of $9$ seconds out of $63$

$\frac{9}{63} = \frac{1}{7} \Rightarrow \mathrm{(D)}$

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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

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