A local detective has a 99% chance of correctly identifying a criminal during a

Question

A local detective has a 99% chance of correctly identifying a criminal during a questioning session and a 0.5% chance of incorrectly classifying an innocent pe criminal during a questioning session. Local police reports indicate that 0.9% of all people have committed a crime. (8pts) What is the probability that a randomly selected person that is brought in for questioning with the detective is classified as a criminal? (4 pts) a. b. If a person who is brought in is determined by the detective to not be a criminal, what is the probability that they are actually not a criminal? (4 pts)

Explanation / Answer

Here we are given that:

P( identified criminal | Criminal ) = 0.99 and P( identified criminal | innocent ) = 0.005

Also we are given that P( criminal ) = 0.009 and therefore P( innocent ) = 1 - 0.009 = 0.991

a) Using law of total probability, we get:

P( identified criminal ) = P( identified criminal | Criminal )P( criminal ) + P( identified criminal | innocent ) P( innocent )

P( identified criminal ) = 0.99*0.009 + 0.005*0.991 = 0.013865

Therefore 0.013865 is the required probability here.

b) Now from previous part, we know that: P( identified criminal ) = 0.013865

Therefore, P( identified innocent ) = 1 - 0.013865 = 0.986135

Using law of bayes theorem, the probability here is computed as:

P( innocent | identified innocent ) = P( identified innocent | innocent ) P( innocent ) / P( identified innocent )

P( innocent | identified innocent ) = 0.995*0.991 / 0.986135 = 0.9999

Therefore 0.9999 is the required probability here.

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