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1999 AMC 8 Problems/Problem 10

Revision as of 18:15, 17 June 2011 by Kingofmath101 (talk | contribs) (1998 AMC 8 #10, Solution)
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Okay, so the whole cycle, $60$ seconds, is our denominator. The light is green for 25 seconds in the cycle, so it is a color other than green for 35 seconds. So, the probability that it will not be green at a random time is $\frac{35}{60}$. This can be simplified, by factoring out the factor of $5$ common to the numerator and the denominator, to $\frac{7}{12}$.

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