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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 Breathalyzer test that will determine whether or not they will be arrested. Trained officers can make the right decision 80% of the time based on the initial brief stop. Suppose the police operate such a checkpoint on a Saturday night after 9pm, a time when national traffic safety experts suspect that about 12% drivers have been drinking. (Draw a tree diagram if needed)

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 driver who was released had actually been drinking?

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Edited Mon, 26 Sep '16
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D = Detained

K = Drinking

Right 80% of the time → P(D|K) = P(D'|K') = 0.8

(It could imply that P(D∩K) + P(D'∩K') = 0.8, but this would lead to insufficient data to solve the problem)

a) P(D|K') = P(D∩K') / P(K') = 0.2 * 0.88 / 0.88 = 0.2

Note, P(D|K') = 1 - P(D'|K') in this case particularly.

b) P(D) = ∑P(D∩$K_i$) = ∑P($K_i$)*P(D|$K_i$) = 0.12 * 0.8 + 0.88 * 0.2 = 0.272

c) P(K|D) = P(K∩D) / P(D) = P(D|K)P(K)/P(D) = 0.8*0.12/0.272 = 0.3529

d) P(K|D') = P(K∩D') / P(D') = P(K∩D') / (∑P($K_i$)P(D'|$K_i$) = 0.12*0.2 / (0.12*0.2 + 0.88*0.8) = 0.024 / 0.728 = 0.03296

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Edited Mon, 26 Sep '16
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