Philadelphia is considering the installation of a camera at a busy intersection
ID: 2955563 • Letter: P
Question
Philadelphia is considering the installation of a camera at a busy intersection to catch people running red lights. Based on prior tickets given by police officers at that intersection, and based on a research study, they estimate that the probability of a car running the light is 0.00007, following a Poisson distribution, with about 50,000 cars passing the intersection each day. For simplicity, ignore the reduction in traffic flow likely to occur on weekends.a. Determine the probability that on any day, no one will run a light.
b. Determine the probability that on any day, more than 2 people will run the light.
c. If tickets cost $100 each, and if about 5% of people receiving tickets will successfully contest their ticket (and not have to pay it), what is the annual projected income if the lights are installed?
d. What factor might change the overall probability from our original estimate, if the cameras are installed? How would this affect projected revenue? How likely is this to occur? (Your estimation, not an exact probability.)
Explanation / Answer
The formula is P(x=k)= (e^(-u) u^k)/k! Where u= mean of the distribution. k= number of successes. Mean of this is 50,000 * .00007=3.5 So the probability of 0 cars is ( e^(-3.5)* 3.5^0)/0!= 0.030 Probability of 1 car is ((e^(-3.5)* 3.5)/1!= 0.106 Probability of 2 cars is ((e^(-3.5)*3.5^2)/2!= 0.185 The probability of more than two cars is 1-0.030-0.106-0.185 = 0.679 Expected revenue would be 365 * 3.5*.95*100=$121,362.50. The cameras might serve as a deterrent to people running red lights, so the average might fall to less than 3.5 per day, reducing yearly revenue.