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Homework: Chapter 13 - Probability Rules! Save ti Score: 0.75 of 1 pt 9 of 10 (10 complete) HW Score: 94.17%, 9.42 of 10 p 13.1.55 Question Help 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 à Breathalyzer test that will determine whether or not they will be arrested. The police say that based on the brief initial stop trained officers can make the right decision 80% of the time. Suppose the police operate a sobriety checkpoint after 9:00 pm. on a Saturday night, a time when traffic safety experts suspect that about 19% of drivers have been drinking. Complete parts a) through d) below. a) A statistics student is stopped at the checkpoint and, of course, has not been drinking. What is the probability that the statistics student is detained for further testing? The probability that the statistics student is detained for further testing is 020 Round to three decimal places as needed.) n b) What is the probability that any given driver will be detained? The probability that any given driver will be detained is 0.314 Round to three decimal places as needed) c) What is the probability that a driver who is detained has actually been drinking? The probability that a driver who is detained has actually been drinking is 0.484 Round to three decimal places as needed ) Question is complete. Tap on the red indicators to see incorrect answers All parts showing Similar Question

Answer 1

a)
probability that student is detained | he has not been drinking
= probability that police makes wrong decision
= 1 - 0.80
=0.20

b)
A - driver is drinking
B - driver is detained
probability that any given driver be detained
P(B) = P(A) P(B|A) + P(A') P(B|A')
= 0.19 * 0.80 + (1-0.19) (1-0.80)
= 0.314

c)
P(A| B)
= P(A and B)/P(B)
= P(A) P(B|A) / P(B)
= 0.19 * 0.80 / 0.314
= 0.484076
= 0.484

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