# Find the probability of his having to stop at the light Answer

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

B) Find his expected delay time due to the light

HallsofIvy
Homework Helper

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