driving

[nbalderaz]

a man drives to work along a route with one traffic light. He drives on a major road, so the light is green three times as long as it is red: specifically, it is green for 45 seconds and red for 15 seconds(ignore yellow). Assume he approaches the light at a random time.

A) Find the probability of his having to stop at the light

Answer .25

B) Find his expected delay time due to the light

Answer 1.875s

[HallsofIvy]

And your question is?

If you have just copied those answers from an answer book and are asking HOW to get those:

Each minute has 4 15 minute sections. For 3 of those, the light is green, for 1, it is red. In terms of "having to stop at the light", there is 1 "success" and 3 "failures" for a total of 4 equally likely outcomes. The probability of a "success" (having to STOP at the light) is the number of "successes" over the number of outcomes: 1/4= 0.25.

Of course, you could have done this in terms of seconds: counting each second in a minute as an outcome, there are 60 equally likely outcomes: 15 of them are "successes" (stopping for the red light) and 45 are "failures": the probability of a "success" is 15/60= 1/4= 0.25

If event "A" has probability p, then "not A" has probability 1-p. The "expected value" of event A is "value of A" * p+ "value of not A" *(1- p).

In this case, "A" is "stops at the red light" and its value is the time delayed: 15 seconds. "not A" is "does not stop at the light" and its value is 0 because the driver is not delayed at all. Since the probability of being stopped is 0.25, the expected value (expected delay time) is 0.25*15 sec+ 0.75*0= 15/4= 3.75 seconds, NOT the 1.875s you give.

1.875= 15/8, not 15/4, half the correct expected time.

[CRGreathouse]

In this case, "A" is "stops at the red light" and its value is the time delayed: 15 seconds. "not A" is "does not stop at the light" and its value is 0 because the driver is not delayed at all. Since the probability of being stopped is 0.25, the expected value (expected delay time) is 0.25*15 sec+ 0.75*0= 15/4= 3.75 seconds, NOT the 1.875s you give.

1.875= 15/8, not 15/4, half the correct expected time.

1.875 is the correct answer. The chance that the driver stops at the light for 15 seconds is infinitesmal; it's more likely that he will arrive some seconds after the light turns red. It's easy to show that the average time stopped at a red light when you stop at all is 7.5 seconds (uniform probability distribution and all), so the average time stopped at a red light is 3/4 * 0 + 1/4 * 7.5 = 1.875.