Drunk Driver Checkpoints
Drunk Driver Checkpoints
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Long Probability problem with 4 questions.?
I am not sure if I am doing the problem correctly. Please, help.
Police sometimes erect sobriety checkpoints. If no suspision is aroused, the driver is released: otherwise, drivers are DETAINED for a breathalyzer test to determine if he is to be arrested. Suppose, that 85% (BOTH for drinking and not-drinking events) cops make the right decision. The Safety Experts say that 12% of drivers have been drinking.
a. If a driver is randomly stopped and hasn't been drinking, what is the probability that he'll be arrested?
b.What's the P that any given driver shall be arrested?
c. What's the P 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?
No need of deep explanations, just to see if i am on the right track.
Thanks
a) 0.15 I believe
b) 0.88 * 0.15 + 0.12 * 0.85
= 0.234
c) 0.12 * 0.85 / 0.234
= 0.435897 (approx)
d) 0.12 * 0.15 / (1 - 0.234)
= 0.023499 (approx)
I'm assuming: 85% make the right decision of detaining
When you say this, it means 85% of drinkers are detained, 85% of non-drinkers aren't.
Being arrested (pt. a, b) is the same as being detained