Conditional probability lie detector

Question

A lie detector returns a positive result when a person is lying in 90% of all cases. Unfortunately, it also results a positive result when a person is telling the truth in 20% of all cases. Statistically, it is estimated that 10% of all test subjects lie. A randomly chosen test subject takes a test and obtains a negative result. What is the probability that the subject told the truth?

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So this was a 2 part question in which I have taken $A=$ Positive Result and $B=$ Subject is Lying: I've then calculated $P(B)=0.1,$ $P(A|B^c)=0.2,$ $P(A|B)=0.9,$ $P(B^c)= 0.9,$ and $P(B|A)=1/3.$ I was thinking of using Bayes' formula and using the law of total probability to figure out $P(B^c|A^c)$ but I am struggling with this. If anyone can point me in the right direction it would be appreciated.

Answer 1

Hint \begin{equation} \begin{array}{c|c|c}&\text{positive}&\text{negative}\\ \hline \text{non liars}&0.18&0.72\\ \hline \text{liars}&0.09&0.01 \end{array} \end{equation}

Answer 2

You may be "struggling" because your answer(s) may be puzzling. [I was trying to avoid giving a final numerical answer, but @Stacker's answer (+1), just now posted, has the same interpretation as mine.]

You are right about using Bayes' Theorem. This problem can be solved with a straightforward application of Bayes' Theorem where the sample space is partitioned into Truth-tellers $(T)$ [with $P(T) = 0.9]$ and Liars $(L = T^c)$ [with $P(L) = .1].$

Denote lie detector results as positive (finds evidence of lying) and negative (no evidence of lying = passes test) by $+$ and $-,$ respectively. Then

$$P(-) = P(T\cap -)+P(L\cap -) = P(T)P(-|T) + P(L)P(-|L),$$ where you can deduce all four probabilities on the right-hand side from the information provided. [For example, $P(-|T) = 1 - P(+|T) = 1-0.2 = 0.8.$ Note the initial sentences that mean 'The lie detector accuses 90% of liars of not telling the truth' and (b) `Unfortunately, the lie detector accuses 20% of truth tellers of being liars.' (This potentially-confusing terminology is also used in medical testing for diseases. A "negative" test means "disease free." A "positive" test means "disease detected.")]

Then $P(T|-) = P(T\cap -)/P(-),$ where you know both probabilities from above.

Note: Unless the lie detector is worse than useless, one would expect $P(T|-) \ge P(T) = 0.9.$

Answer 3

                                                  overall
                 / positive lie detector .9       .09
/ lie         .1    
                 \ negative lie detector .1       .01

                 / positive lie detector .2       .18
\ tell truth  .9
                 \ negative lie detector .8       .72

You could draw a tree, and then get the overall possibility from each branch by multiplying the probabilities along each branch. So you want P(told truth | detector says negative)=$\frac{P(told truth,detector negative)}{P(detectove negative)}=\frac{.72}{.01+.72}=\frac{72}{73}\approx .9863$ by the definition of conditional probability/Bayes formula. So if a negative result is obtained it is highly probable that the person told the truth.