Question

A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 0.027. At the checkpoint, the security system denied a person without security problems 1.5% of the time. Also the security system passed a person with security problems 1% of the time. What is the probability that a person who passed through the system is without any security problems? Report answer to 4 decimal places.

Answer 1

solution:

Let S shows the event that person being a security hazard. So,

P(S) = 0.027

By the complement rule,

p(s^')=1-0.027=0.973

Let D shows the event that security system denied the person and P shows the event that security system passes the person. So we have

P(D|S') = 0.015

P(P|S) = 0.01

By the complement rule we have

P(P|S') = 1 - P(D|S') = 1 - 0.015 = 0.985

By the law of total probability, we have

P(P) = P(P|S)P(S) + P(P|S')P(S') =0.01 * 0.027 + 0.985 * 0.973 =  0.958675

By the Baye's theorem we have

P(S'|P) =[ P(P|S')P(S')] / P(P) = [0.985 * 0.973] / 0.9460 = 0.99971