9.2.3 Geometric Probability (continued)
Example 1 Using Length to Find Geometric Probability
The probabilities of events in time are calculated in this example that uses a length model for geometric probability. The durations of the red, yellow, and green lights for a stoplight are given. A repeating sequence of events in time can be modeled using a line segment. The probability that the light is red at a random arrival to the intersection is equal to the ratio of the time it is red to the total time in one cycle the light goes through. This is equal to the ratio of the length of the line segment representing red to the length of the line segment representing the entire cycle. P = 60/(45 + 5 + 60) = 60/110 ≈ 0.55. The probability of having to wait more then 5 seconds is the probability of arriving while the light is red, except for the last 5 seconds of that period. The probability of this is P = (60 − 5)/110 = 1/2. The expected number of times this will happen out of 80 arrivals is equal to 80 times the probability of it happening, or 80(1/2) = 40 times. This is the expected number of times it will happen, not the actual number, since the event is a chance one like flipping a coin. A length model is used to demonstrate geometric probability in this example. The sample space is given as AD . The lengths of the line segments forming the sample space are given. To find the probability of choosing a point on line segment BD , determine the ratio of the length of BD to the length of the sample space. The length of the sample space, AD , is 5 + 3 + 7 = 15. The length of the event space is 7 + 3 = 10. The probability of the event is 10/15 = 2/3. The probability that the point is not on CD is 1 minus the probability it is on CD . The length of the event space, CD , is 7. The probability P = 1 − (7/17) = 8/15. The probability that the point is on AB or CD is the sum of the individual probabilities: P = 5/15 + 7/15 = 12/15 = 4/5.
Example 2 Transportation Application
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