1. ## Conditional Probability

http://i.imgur.com/9UELFY5.png
Here is the problem in question. I have the answers for both of these already,but I am not sure I have 2b correct.

http://i.imgur.com/9NmUgD8.png

Here are the probability for 2a For 2b I just thought you had to take 95/(95+15) to get the probability, is that right or do I need to include the 13% somehow?

2. ## Re: Conditional Probability

The 13% is relevant to the problem. If you don't use it then you aren't going about the problem correctly.

3. ## Re: Conditional Probability

Originally Posted by Dason
The 13% is relevant to the problem. If you don't use it then you aren't going about the problem correctly.
So, should I organize it something like this? P(drugs)=P(drugs|positive)P(positive)+P(drugs|negative)P(negative) and then rearrange to find P(drugs|positive) ?

4. ## Re: Conditional Probability

That's not the conventional way we would do it but it should give a correct answer. Note that you'll need to figure out P(positive) and P(negative) to go this route.

5. ## Re: Conditional Probability

Originally Posted by Dason
That's not the conventional way we would do it but it should give a correct answer. Note that you'll need to figure out P(positive) and P(negative) to go this route.
Oh there is another route? My lecture notes only have this route. Welp

6. ## Re: Conditional Probability

After doing some searching:
The probability that someone tests positive is (0.95)(0.13) + (0.15)(0.87).
So I want want (0.95)(0.13)/((0.95)(0.13) + (0.15)(0.87)), which gives me 48.62% someone is doping?

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