Question

The government detains a person suspected of a crime, and uses enhanced interrogation techniques to see if the suspect will confess. Suppose that 1 in 100 suspects is truly guilty. Suppose further that, if a suspect is truly guilty, there is a 95% chance that the interrogation will elicit a confession (i.e. they will say that they are guilty), and a 5% chance that the suspect will maintain their innocence (i.e. they will say they are not guilty). If a suspect is truly innocent, there is a 90% chance they will maintain their innocence (i.e. they will say they are not guilty), and a 10% chance they will give a false confession (i.e. they will say they are guilty even though they are innocent). During the interrogation, the suspect confesses. Given this information, what is the probability that this suspect is actually innocent? Round your answer to 2 decimal places.

Answer 1

Solution:

Given:

i) 1 in 100 suspects is truly guilty.

ii) if a suspect is truly guilty, there is a 95% chance that the interrogation will elicit a confession and a 5% chance that the suspect will maintain their innocence

iii) If a suspect is truly innocent, there is a 90% chance they will maintain their innocence and a 10% chance they will give a false confession

Let TG = Truly Guilty ,   TI = Truly Innocent

C = Confesses , I = Innocent

Thus

P(TG) = 1/100 = 0.01 ,

then P( TI) = 1 - P(TG) = 1 - 0.01 = 0.99

P( C | TG ) = 0.95

P( I | TG ) = 0.05

P( I | TI ) = 0.90

P( C | TI) = 0.10

We have to find: P( suspect is actually innocent given that the suspect confesses) = .......?

That is:

P( TI | C ) = .........?

Thus using Bayes rule of probability: