Question

85% of suspects tried for terrorism actually are guilty. (Don’t worry about how we know this, just take
it as a fact for the sake of this problem.) Also, the probability that a person suspected of terrorism will
confess after prolonged questioning is 40% if the suspect is innocent, and 10% if the suspect is guilty.
A suspect, Josef K., is being tried for terrorism. If he confessed after prolonged questioning, what is the
probability that he is actually guilty? And how can you explain this puzzling result? (First of all, you
may need to explain why it is puzzling.

Answer 1

Let \(G\) = guilty
\(I \) = innocent
\(C\) = confesses after prolonged questioning

You are given:
\(P(G)=0.85\), and notice that "innocent, \(I\)" is just the complement of being guilty, so \(P(I)=1-0.85 =0.15\)
\(P(C|I)=0.4\)
\(P(C|G)=0.1\)

You are asked to find:
\(P(G|C)\)

Since \(G\) and \(I\) form a partition (i.e.\( P(G) + P(I)=1\)), you can use Bayes' Theorem:
\[ P(G|C)=\frac{P(C|G)P(G)}{P(C|G)P(G)+P(C|I)P(I)}\] and you have all of those quantities given. Just plug them it.