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You want to find out how many students on campus smoke marijuana. However, you don't want to just go up and ask them if they smoke pot, as they might be embarrassed or lie for some other reason. Therefore, you decide to fill a bag with copies of two essentially the same question:

Question 1: Do you smoke marijuana? Question 2: Do you not smoke marijuana? 

Now, you know the % of Question 1s in your bag (as well as the % of Question 2s in the bag). You go out and ask a bunch of students to grab a question out of the bag, answer it on a separate piece of paper, and put their answers in the answer box.

At the end of the day, you have a full box of answers, and you also know the ratio of Q1s to Q2s that were answered.

Hints: Bayes theorem and set up the experiment such that the ratio of Q1 to Q2 is not 1:1.

The probability of a student on campus being a smoker is P(studentIsSmoker) = something involving P(student said yes)

The question is, what is the "something involving P(student said yes)" formula, and how would you set up the experiment?

Additionally, apparently this could turn out the wrong results, but "probabilistically it will be right"

Seems impossible to me, but there's probably some math magic I don't know about.

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    Discrete and discreet are different words.2012-09-12

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Let $f$ be the fraction of questions that are Question $1$, and let $p$ be the probability that a student is a smoker. If there are $n$ responses altogether, we expect about $fpn$ yes answers from students drawing Question $1$ (why?) and about $(1-f)(1-p)n$ yes answers from students drawing Question $2$. (We expect $(1-f)n$ students to draw Question $2$, and each of those has probability $1-p$ of being a non-smoker and therefore answering yes.) Thus, we expect about $\Big(fp+(1-f)(1-p)\Big)n=(1-f-p+2fp)n$ yes answers altogether. In other words, we expect that the fraction of yes answers will be about $1-f-p+2fp$.

Let $y$ be the actual fraction of yes answers in the sample; $y$ is the observed probability of a student’s saying yes. Then $y$ should be approximately ... what? If you write down the right expression, you can easily solve it for $p$ in terms of the known numbers $f$ and $y$.

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    @Silver: Yep. Looks like I typoed it in the display line and then copied it mechanically.2012-09-12