Answer:
0.9091 = 90.91% probability that a person flagged as a future criminal by the Precogs will actually commit a crime
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
![P(B|A) = \frac{P(B)*P(A|B)}{P(A)}](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28B%29%2AP%28A%7CB%29%7D%7BP%28A%29%7D)
In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
In this question:
Event A: Flagged as future criminal
Event B: Commits a crime.
In the year 2054 the government estimates that out of every 1,000,100 citizens 1,000,000 will remain law-abiding, and 100 will eventually commit a crime.
This means that ![P(B) = \frac{100}{1000100} = 0.00009999](https://tex.z-dn.net/?f=P%28B%29%20%3D%20%5Cfrac%7B100%7D%7B1000100%7D%20%3D%200.00009999)
If a person is going to commit a crime, the Precogs identify that person correctly 99.999% of the time
This means that ![P(A|B) = 0.99999](https://tex.z-dn.net/?f=P%28A%7CB%29%20%3D%200.99999)
Probability of being identified as a criminal:
0.99999 out of 0.00009999(identified and commits crime).
(1-0.99999) out of (1-0.00009999) (identifed and does not commits crime). So
![P(A) = 0.99999*0.00009999 + (1-0.99999)*(1-0.00009999) = 0.000109988](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.99999%2A0.00009999%20%2B%20%281-0.99999%29%2A%281-0.00009999%29%20%3D%200.000109988)
What is the probability that a person flagged as a future criminal by the Precogs will actually commit a crime
![P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.00009999*0.99999}{0.000109988} = 0.9091](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28B%29%2AP%28A%7CB%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.00009999%2A0.99999%7D%7B0.000109988%7D%20%3D%200.9091)
0.9091 = 90.91% probability that a person flagged as a future criminal by the Precogs will actually commit a crime