Question

2. Consider the following problem associated with synchronizing traffic lights. A particular traffic light (Light A) has a cycle as follows: Red = 1 min Green = 1.5 min Yellow = 0.5 min Some distance before Light A is another light (Light B). Due to varying drivers and conditions, the travel time between the two lights varies from vehicle to vehicle. Data suggest that 40% of all cars leaving Light B at a time when Light A is red are not delayed by Light A when they reach it, 80% of all cars leaving Light B during a green cycle at Light A are not delayed, and 20% of all cars leaving during a yellow cycle are not delayed. a) What is the probability that a car leaving Light B during a red cycle at Light A is delayed when it reaches Light A? b) Compute the total probability (regardless of cycle) that a car is delayed. c) Given that a car was delayed by Light A, what is the probability that it left Light B while Light A was (i) red? (ii) green? (iii) yellow?

Answer 1

Let R,G and Y denote that light A is red, green and yellow respectively when the car left light B

And D be the event that the car was delayed by Light A

Thus, P(R) = 1/(1 + 1.5 + 0.5) = 1/3

P(G) = 1.5/3 = 1/2

P(Y) = 0.5/3 = 1/6

P(D'|R) = 0.4, P(D'|G) = 0.8, P(D'|Y) = 0.2

(a) The required probability = P(D|R) = 1 - P(D'|R) = 0.6

(b) The required probability = P(D)

= P(D|R)*P(R) + P(D|G)*P(G) + P(D|Y)*P(Y)

= 0.6*1/3 + 0.2*1/2 + 0.8*1/6

= 13/30

(c)

(i) P(R|D) = P(D|R)*P(R)/P(D) = (0.6*1/3)/(13/30) = 6/13

(ii) P(G|D) = (0.2*1/2)/(13/30) = 3/13

(iii) P(Y|D) = (0.8*1/6)(13/30) = 4/13