A lie detector will be consulted in a court. It is known that the detector gives the correct result (guilty) for a guilty suspect $90\%$ of the time and it gives the result not guilty for an innocent suspect $99\%$ of the time. From statistics of the tax authorities, it is known that $5\%$ of the citizens cheat seriously in their tax returns . The lie detector indicates that the person is guilty of cheating. What is the probability that the suspect is actually innocent?
$T =$ Tells the truth
$TT =$ Detector thinks it is truth
So I got now, that the probability that someone is telling truth given that the machine thinks is truth, $P(T\mid TT) = $ is around $1881/1891$. But what I need to know, is the probability that someone is actually telling the truth given that the machine thinks it is a lie: $P(T\mid \neg TT)$.
I would simply say that would be $10/1891$, but how can I prove this?
By Bayes rule (and the law of total probability for the denominator) \begin{align}P(T\mid \neg TT)&=\frac{P(\neg TT\mid T)P(T)}{P(\neg TT)}\\[0.4cm]&=\frac{P(\neg TT\mid T)P(T)}{P(\neg TT\mid T)P(T)+P(\neg TT\mid \neg T)P(\neg T)}\\[0.4cm]&=\frac{\frac{1}{100}\cdot0.95}{\frac{1}{100}\cdot 0.95+\frac{90}{100}\cdot0.05}\approx 0.174\end{align}