Question

A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the checkpoint, the security system denied a person without security problems 2% of the time. But the security system passed a person with security problems 1% 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?

Answer 1

(a) 0.9412

(b) 0.9996 ≈ 1

Explanation:

Denote the events a follows:

`P` = a person passes the security system

`H` = a person is a security hazard

Given:

`P (H) = 0.04,\ P(P^(c)|H^(c))=0.02\ and\ P(P|H)=0.01`

Then,

`P(H^(c))=1-P(H)=1-0.04=0.96\\P(P|H^(c))=1-P(P|H)=1-0.02=0.98\\`

(a)

Compute the probability that a person passes the security system using the total probability rule as follows:

The total probability rule states that:
`P(A)=P(A|B)P(B)+P(A|B^(c))P(B^(c))`

The value of P (P) is:

`P(P)=P(P|H)P(H)+P(P|H^(c))P(H^(c))\\=(0.01*0.04)+(0.98*0.96)\\=0.9412`

Thus, the probability that a person passes the security system is 0.9412.

(b)

Compute the probability that a person who passes through the system is without any security problems as follows:

`P(H^(c)|P)=(P(P|H^(c))P(H^(c)))/(P(P)) \\=(0.98*0.96)/(0.9412) \\=0.9996\\\approx1`

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.