Question

A biometric security device using fingerprints erroneously refuse to admit 1 in 1,000 authorized person from a facility containing classified information. The device will erroneously admit 1 in 1,000,000 unauthorized persons. Assume that 95 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?

Answer 1

0.01864

Explanation:

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)

putting value we get

P(R) = P(A)P(R|A) + P(NA)P(R|NA) = 0.95*(1/1000) + 0.05*((1,000,000-1)/1,000,000) = 0.0.05094995

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

= [0.95*(1/1000)]/(0.05094995)

=0.01864