Police often set up sobriety checkpoints--roadblocks where drivers are asked a f
ID: 2960041 • Letter: P
Question
Police often set up sobriety checkpoints--roadblocks where drivers are asked a few brief questions to allow the officer to judge whether or not the person may have been drinking. If the officer does not suspect a problem, drivers are released to go on their way. Otherwise, drivers are detained for a Brethalyzer test that will determine whether or not they will be arrested. The police say that based on the brief intial stop, trained officers can make the right decision 80% of the time. Suppose the police operate a sobriety checkpoint after 9p.m. on a Saturday night, a time when national traffic experts suspect that 12% of drivers have been drinking.
a) You are stopped at the checkpoint and, of course, have not been drinking. What's the probability that you are detained for further testing?
b) What's the probability that any given driver will be detained?
c) What's the probability that a driver who is detained has actually been drinking?
d) What's the probability that a driver who was released had actually been drinking?
Explanation / Answer
A=drinking B=detained Ac=complement of A-->not drinking Bc=complement of B-->not detained a) P(B/Ac)=0.20 The police are incorrect 20% of the time if they are right 80% ofthe time. b) Two chances to be detained: not drinking and the police were mistaken or drinking and police were correct. P(any detained)= (0.80x0.12) + (0.88x0.20) =0.272 or 27.2% chance any given driver will be detained. c) Use Baye's Theorem: P(A/B)=P(A)P(B/A) divided by P(A)P(B/A)+P(Ac)P(B/Ac) =(.012x0.80) divided by (0.12x0.80)+(0.88x0.20) =(0.096) divded by (0.272) =0.353 or 35.3% chance detained driver had actually been drinking. d) Bc=not detained (or released) P(A/Bc=(0.12x0.20) divided by (0.12x0.80)+(0.88x0.8) =(0.024) divided by (0.728) =0.033 or 3.3% chance of being released and drinking