A biometric security device using fingerprints erroneously refuses to admit 3 in

Question

A biometric security device using fingerprints erroneously refuses to admit 3 in 1,500 authorized persons from a facility containing classified information. The device will erroneously admit 3 in 1,005,000 unauthorized persons. Assume that 98 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?

A biometric security device using fingerprints erroneously refuses to admit 3 in 1,500 authorized persons from a facility containing classified information. The device will erroneously admit 3 in 1,005,000 unauthorized persons. Assume that 98 percent of those who seek access are authorized.

Explanation / Answer

P(authorised) = 0.98

P(not authorised) = 1 - 0.98 = 0.02

By theorem of total probability,

P(refused admission)

= P(refused admission | authorised) * P(authorised) + P(refused admission | not authorised) * P(not authorised)

= (3/1500)*0.98 + (1 - 3/1005000)*0.02 = 0.02196

By Bayes theorem,

P(authorised | refused admission) = P(refused admission | authorised) * P(authorised) / P(refused admission)

= (3/1500)*0.98 / 0.02196 = 0.0892

= 0.999969541

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