Question

In a certain city, violent criminals make up 0.2% of the population. Professional profiler, Mary Mac, states that 99% of the violent criminals match the profile she has created. City officials are considering the following policy: "Let's preemptively throw all those people who match the profile into a detention/rehab center." During discussions, the police chief makes the fallacious statement, "Ms Mac tells us that 99% of those who match her profile are violent criminals, so I am strongly in favor of the preemptive detention policy." To illustrate the consequences of the fallacy, suppose that 5% of those who are not violent criminals match Ms Mac's profile. If the preemptive detention policy were adopted, what fraction of those thrown in the detention center would not be violent criminals?

A. 1%

B. 5%

C. 96.2%

Answer 1

Let A denote the event that the profile matches as a violent criminal.

Let B denote the event of being a violent criminal

Now, P(B) = 0.2% = 0.002, thus P(B') = 0.998

P(A|B) = 99% = 0.99

P(A|B') = 5% = 0.05

Thus, From law of total probability , P(A) = P(A|B)*P(B) + P(A|B')*P(B') = 0.99*0.002 + 0.05*0.998 = 0.05188

Now, to calculate P(B'|A) = P(A|B')*P(B')/P(A) [ From Nate's theorem]

= 0.05*0.998/0.05188 = 0.962 = 96.2%

Thus, the correct answer is option (C)