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wiki gives this Drug testing Example to illustrate Bayes' theorem

${\displaystyle {\begin{aligned}P({\text{User}}\mid {\text{+}})&={\frac {P({\text{+}}\mid {\text{User}})P({\text{User}})}{P(+)}}\\&={\frac {P({\text{+}}\mid {\text{User}})P({\text{User}})}{P({\text{+}}\mid {\text{User}})P({\text{User}})+P({\text{+}}\mid {\text{Non-user}})P({\text{Non-user}})}}\\[8pt]&={\frac {0.99\times 0.005}{0.99\times 0.005+0.01\times 0.995}}\\[8pt]&\approx 33.2\%\end{aligned}}} $

where, the test will produce 99% true positive results for drug users and 99% true negative results for non-drug users. 0.5% of people are users of the drug.

Note:

wiki does not give 0.01 directly.

how to get the false positive $P({\text{+}}\mid {\text{Non-user}}) = 0.01$ in this case?

"1 - true negative" or "1 - true positive"?

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  • $\begingroup$ The terminologies 'false positive', 'true negative', etc. have become so corrupted by misuse as to have become almost useless. For example, some people, say the probability of 'false positive' is P(Test + and Non-User) and some say it's P(Test + | Non-user). // However, @User158565 gives a correct answer (+1). The terminology 'Specificity of the test' for P(Test - | Non-user) is less abused and still useful. $\endgroup$
    – BruceET
    Commented Aug 1, 2019 at 0:16
  • $\begingroup$ @BruceET would you please point out which part misused the terminologies in my post? $\endgroup$
    – czlsws
    Commented Aug 1, 2019 at 0:54
  • $\begingroup$ I'm not saying that you misused 'true negative' and 'true positive'. I'm suggesting that more precise terminology (in terms of conditional probabilities) be used because of massive misuse of terminology in some textbooks and many online posts. $\endgroup$
    – BruceET
    Commented Aug 1, 2019 at 2:23

2 Answers 2

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Actually, you are calculating the false positive rate (FPR) instead of the false positive, it seems that wiki example uses this term in a non-rigorous way.

Usually, FP represents false positive, TN represents true negative, and false positive rate

${\displaystyle \mathrm {FPR} ={\frac {\mathrm {FP} }{\mathrm {N} }}={\frac {\mathrm {FP} }{\mathrm {FP} +\mathrm {TN} }}=1-\mathrm {TNR} }$,

where true negative rate ${\displaystyle \mathrm {TNR} ={\frac {\mathrm {TN} }{\mathrm {N} }}={\frac {\mathrm {TN} }{\mathrm {TN} +\mathrm {FP} }}=1-\mathrm {FPR} }$

is provided by your example, which is 99%.

So, the FPR, which you are actually asking for, is equal to 1 - 99% = 1%.

The formulas above comes from this wiki page where you may give more info.

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"99% true negative results" = $P({\text{-}}\mid {\text{Non-user}})$

So $P({\text{+}}\mid {\text{Non-user}}) = 1 - P({\text{-}}\mid {\text{Non-user}}) 1 - 0.99 = 0.01$

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