A new security system needs to be evaluated at a federal building. The probabili

Question

A new security system needs to be evaluated at a federal building. The probability of a person being a security hazard is 1%. At the entrance, the security system allowed a person without any security problems 97% of the time. The system also allowed a person with security problems 5% of the time.

a. What is the probability that a person passes through the system?

b. What is the probability that a person who passes through the system is without any security problems?

Please help me to understand this question!

Explanation / Answer

P(Security hazards) = 0.01; So P(No Security hazards) = 1 - 0.01 = 0.99

P(Pass|No security hazard) = 0.97; So, P(Not pass| No security hazard) = 1 - 0.97 = 0.03

P(Pass|Security hazard) = 0.05; So, P(Not pass| Security hazard) = 1 - 0.05 = 0.95

Hence,

a) P(Person passes through the system)

= P(Pass|No security hazard)* P(No security hazard) + P(Pass|Security hazard)*P(Security hazard)

= 0.97*0.99 + 0.05*0.01

= 0.9608

b) P(No security problem | Pass)

= P(No security problem and Pass)/ P(Pass)

= (0.97*0.99)/ 0.9608

= 0.9995 [Rounded off to 4 decimal places]

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