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Potts has invented a new lie-detector machine and is testing it. One hundred test subjects are told to lie, and the machine catches 80 of them in the lie. Another hundred test subjects are told to tell the truth, but the machine nevertheless thinks that 5 of them are lying.

Let’s say the local police department begins to use the machine in interrogations, and suppose on average 15% of people arrested lie in their interrogations. When the machine indicates a lie, what is the probability that the suspect is really lying? If the machine does not indicate a lie, what is the probability that the suspect is really telling the truth?

My solution if the probability that a test subject is chosen at random from the 200 in fact lie = 80/(80+5)=0.94

Then when 15% of people arrested lie, the probability that the suspect is really lying is 0.15*0.94=0.141

Is this correct?

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  • $\begingroup$ Which 200 do you say a subject is chosen at random from? Your subject should be chosen at random such that there's a 15% chance that he actually lies; if you choose from a pool with 100 liars and 100 truth-tellers, that probablity would be 50% rather than the specified 15%. $\endgroup$ – hmakholm left over Monica Mar 14 '17 at 18:43
  • $\begingroup$ (Oh, and by the way, lie detectors are complete bunk, and the proposed testing methodology is inane. Being told to lie in a test is a completely different experience than lying to the police to save your skin, and being told to tell the truth in a safe, non-threating situation is a completely different experience than being interrogated by police, even as an actual innocent, in a situation where a misstep, malfunction or mis-measured sweat drop could cost you your freedom. The test scenarios have essentially nothing to do with the production use cases). $\endgroup$ – hmakholm left over Monica Mar 14 '17 at 18:47
  • $\begingroup$ The 200 given in 1st paragraph (100 told to lie & 100 told to say truth). Actually, 2nd para is also question. So, if 15% of people detected lying, what is the probability that the suspect is really lying? $\endgroup$ – New_Coder Mar 14 '17 at 18:48
  • $\begingroup$ @New_Coder the 200 test subjects in the first paragraph are not the people arrested in the second paragraph. $\endgroup$ – Graham Kemp Mar 15 '17 at 0:00
  • $\begingroup$ No they are not. Have a look: faculty.cord.edu/ahendric/2008Fall203/BayesWS.pdf $\endgroup$ – New_Coder Mar 15 '17 at 0:12
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No, it is not correct. In fact, I am not sure what that first number you calculate even means ... You say you want to calculate the probability that a person lies ... But we already know that is $0.15$.

Using:

$T$: person teels truth

$TT$ Lie detector thinks person is telling truth

$P(TT) = P(TT|T)*P(T)+P(TT|\neg T)*P(\neg T)= 0.95*0.85+ 0.2*0.15$

Using Bayes's theorem:

$$P(T|TT) = \frac{P(TT|T)*P(T)}{P(TT)} =$$

$$\frac{0.95*0.85}{0.95*0.85+0.2*0.15}$$

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  • $\begingroup$ Please have a look at: faculty.cord.edu/ahendric/2008Fall203/BayesWS.pdf I need to answer 1(b, c, d, e) $\endgroup$ – New_Coder Mar 14 '17 at 18:59
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    $\begingroup$ @New_Coder Ah, ok, I see now what you were trying to do. OK, the answer I provided is for d), and for e) you should of course use the same formula, but plug in some different numbers as indicated. OK, for b) you know for 85 of the 200 subjects the machine thinks that they lie. Since 80 of those 85 were actually lying, the chance you are dealing with any of them is $\frac{80}{85}$. For c), follow a similar recipe. $\endgroup$ – Bram28 Mar 14 '17 at 19:23
  • $\begingroup$ I tried to solve all of them, please have a look: (b) 80/(80+5)=0.9412. (c) 95/(95+20)=0.8261. (d) When the machine indicates a lie: (0.80*0.15)/((0.80*0.15)+(0.05*0.85))=0.7385 When machine does not indicate a lie: (0.95*0.85)/((0.95*0.85)+(0.2*0.15))=0.9642 (e) (0.80*0.03)/((0.80*0.03)+(0.05*0.97))=0.331. $\endgroup$ – New_Coder Mar 14 '17 at 23:26
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    $\begingroup$ @New_Coder Looks good!! $\endgroup$ – Bram28 Mar 14 '17 at 23:29

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