17) At a courthouse every person visiting must pass through an explosives detect

Question

17) At a courthouse every person visiting must pass through an explosives detector. The explosives detector is 90% accurate when detecting the presence of explosives on a person but suffers from a 5% false positive rate. Past studies have determined that the probability that a random person will bring explosives into the courthouse is 0.1%. If the detector indicates that a random person has concealed explosives, what is the true probability they have explosives? FP,05% .40 7 ( b.)0.0177 a.) 0.0509 c.) 0.0009 d.) 0.9000 e.) None of these. ..ol

Explanation / Answer

Let H be the probability that a person has explosive, and H' the probability of not having explosives.

Let D be the probability that the device detects explosives.

Given P(H) = 0.1% = 0.001 and P(H') = 1 - 0.001 = 0.999

Given that probability of detecting given that a person has explosives = P(D/H) = 0.9 and

a false positive, which is probability of detecting when a person does not have explosives = P(D/H') = 0.05

To find the probability that given a detection of explosives, the person has explosives = P(H/D)

By Bayes Theorem: P(H/D) = P(D/H) * P(H)/ P(D)

The Probability of detection P(D) = P(D/H) * P(H) + P(D/H') * P(H')

= (0.9 * 0.001) + (0.05 * 0.999) = 0.0009 + 0.04995

Therefore P(D) = 0.05085

Therefore P(H/D) = 0.9 * 0.001/ 0.05085 = 0.0177

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