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Problem:
A person drives to work via a road with a single traffic signal. The light cycles, green for 45 seconds, red for 15 seconds – ignore yellow. Assume the person approaches the signal at a random time. Please find:
a. the probability of having to stop at the light;
b. the expected delay time due to the signal.

Attempt at Solution:
I was able to deduce that given the total 60 seconds, the probability of the light being green is 75% and the probability of the light being red is 25%. This, however, is where I am stuck.

Swamp G
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1 Answers1

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So I imagine you concluded that the probability of having to stop is $\frac{1}{4}$.

Now to the expected delay. If the light is green, the expected delay is $0$. If the light is red, by symmetry the average delay is $7.5$ seconds. Thus the average delay, in seconds, is $$(3/4)(0)+(1/4)(7.5).$$

André Nicolas
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  • What exactly do you mean by "by symmetry"? – Swamp G Jul 22 '14 at 22:35
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    I am assuming that the time of arrival is uniformly distributed in the $15$ second interval that red is on, indeed in the minute that the light cycles through. That implies that the mean wait is $15/2$. The uniform distribution in the interval $[0,15]$ has mean $7.5$, since the density function is symmetric about $x=7.5$. – André Nicolas Jul 22 '14 at 22:38