# Difference between revisions of "1999 AMC 8 Problems/Problem 10"

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− | + | ==problem== | |

+ | |||

+ | A complete cycle of a tra±c light takes 60 seconds. During each cycle the light is | ||

+ | green for 25 seconds, yellow for 5 seconds, and red for 30 seconds. At a randomly | ||

+ | chosen time, what is the probability that the light will NOT be green? | ||

+ | (A) | ||

+ | 1/4 | ||

+ | (B) | ||

+ | 1/3 | ||

+ | (C) | ||

+ | 5/12 | ||

+ | (D) | ||

+ | 1/2 | ||

+ | (E) | ||

+ | 7/12 | ||

+ | |||

+ | ==solution== | ||

+ | |||

+ | (E) 7/12: | ||

+ | time not green/total time | ||

+ | = | ||

+ | (R + Y)/(R + Y + G) | ||

+ | = | ||

+ | (35/60) | ||

+ | = | ||

+ | (7/12) | ||

+ | : | ||

+ | |||

+ | OR | ||

+ | |||

+ | The probability of green is | ||

+ | (25/60) | ||

+ | = | ||

+ | (5/12) | ||

+ | So the probability of not green is 1- (5/12) = | ||

+ | (7/12) | ||

+ | . |

## Revision as of 15:09, 4 November 2012

## problem

A complete cycle of a tra±c light takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds, and red for 30 seconds. At a randomly chosen time, what is the probability that the light will NOT be green? (A) 1/4 (B) 1/3 (C) 5/12 (D) 1/2 (E) 7/12

## solution

(E) 7/12: time not green/total time = (R + Y)/(R + Y + G) = (35/60) = (7/12)

OR

The probability of green is (25/60) = (5/12) So the probability of not green is 1- (5/12) = (7/12) .