Anew security system needs to be evaluated in the airport. the probability of a person being a security hazard is 0.02. 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 random person does not pass through the system and is without any security problems? report answer to 3 decimal places.

The probability that a random person does not pass through the system and is without any security problems is 0.965

Step-by-step explanation:

We are given that  The probability of a person being a security hazard is 0.02.

P(security) = 0.02

So, P(Without security)=1-P(security)  = 1- 0.02=0.98

We are also given that the security system denied a person without security problems 1.5% of the time

So, P(Not Pass | Without security) = 0.015

Now we are given that  the security system passed a person with security problems 1% of the time

So,P(Pass| With security)=0.01

P(Pass | Without security ) = 1-P(Not Pass | Without security)= 1-0.015 = 0.985

Now we are supposed to find the probability that a random person does not pass through the system and is without any security problems

P(Pass | With security) = P(Pass | Without security ) * P(Without security) + P(Pass | With security ) * P(Security)

P(Pass | With security) = 0.985 *0.98+ 0.01 * 0.02

So, P(Pass | With security) = 0.9655

Hence the probability that a random person does not pass through the system and is without any security problems is 0.965

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)