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

she will most likely have 33  

step-by-step explanation:

75

step-by-step explanation:

you do x-29=46 (the number of cantaloupes sold)

in order to find x, you have to find what minus 29 equals 46, meaning you'd have to add 29 to 46.

29+46=75

if you replace x with 75, the equation looks like:

75-29=46

x=75