<p>Hi there everyone
I am self studying Stats and for some reason I cannot figure this problem out for the life of me (though I'm pretty sure it's an easy one)
The problem concerns probability, etc</p>
<p>Basically the problem states </p>
<p>Police officers have check points where they check to see if drivers have been drinking by asking them some questions. "Trained officers can make the right decision 80% of the time (in regards to whether a person should be held for more questioning or let go). The officers set up a checkpoint on a Saturday night at 9 pm, a time where about 12% of the drivers have actually been drinking. </p>
<p>a)You are stopped at the checkpoint, and you have not been drinking. What's the probability that you are detained for further testing
-I managed to figure out that the answer to this part is .20 or 20%
b)Whats the probability that any given driver will be detained?
c)whats the probability that a driver who is detained has actually been drinking?
d)whats the probability that a driver who was released had actually been drinking</p>
<p>As I said before, I honestly do not even know what to do on parts b-d (I have a feeling it might have something to do with Bayes Rule maybe??)
Please Help
thanks</p>
<p>It’s been a long time since we’ve done this, but i’ll try:</p>
<p>B. There are two ways for a driver to get detained. They either were NOT drinking and got detained, or they WERE drinking and got detained. We also have to consider the fact that the proportion of drivers who’ve been drinking is only 0.12, so that has to be factored into the calculations for drivers who WERE drinking.</p>
<p>P(not drinking and detained) = (0.20) = 0.024.
P(drinking and detained) = (0.12)(0.80) = .096</p>
<p>(.024) + (.096) = .12</p>
<p>Therefore there is a 12% chance that a driver at random would be detained.</p>
<p>C. A driver who has been detained has two outcomes: not drinking (.20 chance) and drinking (.80 chance). Therefore P(detained and drinking)=.80 </p>
<p>D. The probability of getting detained is .88 (1-0.12) and the probability of drinking is 0.12, so (0.88) X (0.12) = 0.1059</p>
<p>P(detained and drinking) = 0.1059</p>
<p>I probably got these all wrong, I wouldn’t be the least bit surprised. But I think I had the general idea. We did that way back in November/December.</p>
<p>BTW… you’re really behind if you’re only on conditional probability.</p>
<p>hahah yes its pretty bland to say the least
im doing kinda ok self studying, but I heard the teacher who teaches it at our school suckss bad. Its kinda funny cause people are asking me for help instead of the teacher
Good luck to you too man-hope ur able to get through it lol</p>