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

Answer:

(a) 0.9412

(b) 0.9996 ≈ 1

Step-by-step 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:

$P(P)=P(P|H)P(H)+P(P|H^{c})P(H^{c})\\=(0.01\times0.04)+(0.98\times0.96)\\=0.9412$

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:

$P(H^{c}|P)=\frac{P(P|H^{c})P(H^{c})}{P(P)} \\=\frac{0.98\times0.96}{0.9412} \\=0.9996\\\approx1$

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