Question

The following case study was reported in the article "Parking Tickets and Missing Women," which appeared in an early edition of the book Statistics: A Guide to the \(U n\) known. In a Swedish trial on a charge of overtime parking, a police officer testified that he had noted the position of the two air valves on the tires of a parked car: To the closest hour, one was at the one o'clock position and the other was at the six o'clock position. After the allowable time for parking in that zone had passed, the policeman returned, noted that the valves were in the same position, and ticketed the car. The owner of the car claimed that he had left the parking place in time and had returned later. The valves just happened by chance to be in the same positions. An "expert" witness computed the probability of this occurring as \((1 / 12)(1 / 12)=1 / 44\). a. What reasoning did the expert use to arrive at the probability of \(1 / 44\) ? b. Can you spot the error in the reasoning that leads to the stated probability of \(1 / 44\) ? What effect does this error have on the probability of occurrence? Do you think that \(1 / 44\) is larger or smaller than the correct probability of occurrence?

Short answer

The expert assumed the valve positions to be independent and hence calculated the combined probability as \(1 / 44\). This is incorrect because once the position of one valve is established, the position of another is not uniformly distributed. Also, a calculation error was made in the answer as it should be \(1 / 144\). The correct probability considering both valves' positions at the same time should be \(1 / 12\).

Step-by-step solution

Identify expert's reasoning

The expert concluded that the probability of the car owner leaving and returning to the same parking spot with the tire valves in the exact same position as \(1 / 44\) by assuming each hour on a clock (like a 12-hour clock) as a possible position for the valves. Hence, with two valves, the expert considered the probability of one valve being in a particular position as \(1 / 12\) and the probability of the second valve being in the same position as again \(1 / 12\). The combined probability is the product: \((1 / 12) * (1 / 12) = 1 / 144\).

Spot the error

The expert made a mistake by considering the position of each valve as independent of one another. Actually, once the position of the first valve is established, the position of the second one is no longer uniformly distributed. Moreover, the probability was miscalculated as \(1 / 44\), whereas the calculations give \(1 / 144\).

Correct Probability

A correct approach would have been to consider the two positions as one event. Meaning, the correct statement is 'What is the probability that in any random hour, both air valves point to exactly 1 o'clock and 6 o'clock position respectively?'. As there are 12 hours on a clock, the correct probability of such an event would be \(1 / 12\).