Given the following situation: 99.9 percent of people have read a certain book. 99.7 percent of the people, who have read the book, can answer a certain question about it. Of these, who did not read the book, 0.04% answer the question correctly by chance. How big is the probability, that you have read the book, if you answered wrong?

My approach: 99.9%*0.3% have read the book and answered wrong. 0.1%*0.96% have not read the book and answered wrong. In total: 99.9%*0.3%+0.1%*0.96% answered wrong.

So the probability should be: (99.9%*0.3%) / (99.9%*0.3%+0.1%*0.96%)

which is the percentage of people, who have read the book and answered wrong divided by the percentage of people who answered wrong. Is that correct?


1 Answer 1


Yes.   Well, almost.   Check your numbers. $\mathsf P(\textrm{Read}\mid \textrm{Not-Correct}) \\= \dfrac{\mathsf P(\textrm{Read})\,\mathsf P(\textrm{Not-Correct}\mid \textrm{Read})}{\mathsf P(\textrm{Read})\,\mathsf P(\textrm{Not-Correct}\mid \textrm{Read})+\mathsf P(\textrm{Not-Read})\,\mathsf P(\textrm{Not-Correct}\mid \textrm{Not-Read})}\\ = \dfrac{99.9\%\cdot 0.3\% }{ 99.9\%\cdot 0.3\%+0.1\%\cdot \color{red}{99.96}\% } \\ \approx 74.9\%$

  • $\begingroup$ Thank you, this is what I got, too. This is kind of a quiz, unfortunately the answer is not accepted for some reason. Do you see another interpretation of the text? $\endgroup$
    – Ctx
    Dec 8, 2015 at 9:52
  • $\begingroup$ Well, I get the same analysis, but the numerical answer rounds to $74.989$ percent $\to 75.0$ percent. Is that possibly a problem? $\endgroup$
    – Brian Tung
    Dec 8, 2015 at 18:25

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