A biometric security device using fingerprints erroneously refuses to admit 2 in 1,900 authorized persons from a facility containing classified information. The device will erroneously admit 1 in 1,019,000 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? (Round your answer to 4 decimal places.)

Using Bayes Theorem: P(authorized | alarm) = P(authorized & alarm)/P(alarm)

P(authorized & alarm) = 0.95 x 2/1900
P(alarm) = 0.95*2/1900 + 0.05*1018999/1019000

Can you take it from here?

95% of the people seeking admission are authorized, but the device outputs a false negative 2 out of 1900 times. Hence the probability that a given person will be authorized and the alarm will be triggered is 0.95 * (2/1900) = 0.001. In other words, out of every 1000 people seek admission, we can expect that 1 will be refused access despite the fact that they're supposed to be authorized. Note that this is NOT the answer to your question!!

Meanwhile, 5% of the people seeking admission are not authorized, and the alarm will trigger 1,018,999 out of 1,019,000 times that one of them tries. Hence, the probability that a person is unauthorized and an alarm will sound is 0.05 * (1,018,999/1,019,000) = 0.04999995 or just barely less than 50 times out of 1000.

Thus, in general, the total probability of the alarm sounding is (.04999995 + .001) = .05099995 or about 51 times out of 1000.

Now, since you're given that the alarm has sounded, the probability that the person seeking admission is authorized is equal to 0.001 / 0.05099995 = 0.0196. So the chances are about 1 in 51.

Note that once you've calculated the total probability of the alarm going off, you can do this with Bayes Theorem. A represents a person being authorized, and B represents the alarm going off:

Woops, you beat me to it EB!

I was reading a book last week called "The Drunkard's Walk" by Leonard Mlodinow (about probability and randomness). In the book he recounted a scary incident from the late 80's which was analogous to this question. After a routine blood test to increase his life insurance, he was denied coverage. Upon querying for more of an explanation, he received a message from the doctor who took the blood, which said that it turned out he'd gotten a positive result on an HIV test! The doctor told him there was a 99.9% probability that he was HIV-positive. This was late on a Friday, and he had to spend the entire weekend with that horrific news hanging over his head. Upon speaking to the doctor in more detail when he returned the following Monday, it turned out that the false-negative rate for the test was 0.1%, so the doctor assumed that meant that there was a 99.9% chance that he was positive. After somehow managing not to throttle the coctor, the author proceeded to explain the massive difference between the probability the doctor had quoted and the conditional probability that arises from a positive result. He explained to the doctor that since the infection rate among monogamous, heterosexual, non-intravenous drug using men was around 1 in 10,000, that meant that there was really only about a 10% chance of him being positive. A second test confirmed that it was, indeed, a false positive.