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 14:09, 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) .
