A biometric security device using fingerprints erroneously r

A biometric security device using fingerprints erroneously refuses to admit 1 in 1,800 authorized persons from a facility containing classified information. The device will erroneously admit 1 in 1,015,000 unauthorized persons. Assume that 90 percent of those who seek access are authorized. If the alarm goes off and a person is refused admission, what is the probability that the person was really authorized? (Round your answer to 5 decimal places.)

Solution

Let A be the event that a person was authorized and NA be the event that the person was not authorized. Let R be the event that a person was refused admission and G be the event that a person was granted admission.

Then we have to find P(A|R) = [P(A)*P(R|A)]/P(R)

P(R) = P(A)P(R|A) + P(NA)P(R|NA) = 0.9*(2/1000) + 0.1*((1001000-2)/1001000) = 0.1018

P(A|R) = [P(A)*P(R|A)]/P(R)

= [0.9*(2/1000)]/(0.1018)

= 0.01768