I saw this question which I found confusing: Suppose 400 people are surveyed. For each person, they flip a coin and if it lands heads they are asked,

A: Are you a student?

otherwise they are asked

B: Do you shoplift?

230 of the 400 surveyed answered yes. Of the students, estimate the proportion who shoplift.

Personally I would attempt to solve this with conditional probabilities, but it seems to me there is inadequate information for that. Here's my attempt.

$$ \frac{230}{400} = P(Yes)$$

$$ P(Shoplift|Student) = P(Yes|Tails)$$

$$P(Yes) = P(Yes | Tails)P(Tails) + P(Yes|Heads)P(Heads)$$

$$P(Tails)=P(Heads) = .5$$

Other than this I don't think any information is given and useful, and this amount of information seems inadequate to solve for $P(Shoplift|Student)$. The most you can infer is

$$\frac{230}{400} = (.5)(P(Yes|Tails) + P(Yes|Heads)) \Rightarrow$$

$$P(Yes|Tails) = \frac{23}{20} - P(Yes|Heads)$$


Independent estimates of proportions (which is what the coin toss is achieving) means you cannot improve or even modify the conditional probability in students beyond what is provided by the unconditional estimate. Despite your question text, you do not actually have any "cross-tabulated" information.

  • $\begingroup$ About cross-tabulation, fair enough, I am pretty sure I get what you mean: I don't actually have the data for any cell in the two-by-two grid of frequencies. But the first part, I'm not super clear on what you're saying, though it sort of sounds like you're saying "Yes, the given information is insufficient." $\endgroup$
    – Addem
    Mar 7 '16 at 3:34
  • $\begingroup$ Yes, the information is insufficient. $\endgroup$
    – DWin
    Mar 7 '16 at 4:58

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