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A new 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.

User Suroh
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2 Answers

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Final answer:

Using the given probabilities, we find that the probability is 0.965, or 96.5%.

Step-by-step explanation:

To find the probability that a random person does not pass through the system and is without any security problems, we need to calculate the complement of two events: a person being a security hazard and the security system denying a person without security problems.

First, let's calculate the probability of a person being a security hazard:

Probability of a person being a security hazard = 0.02

Next, let's calculate the probability of the security system denying a person without security problems:

Probability of the security system denying a person without security problems = 1.5% = 0.015

To find the probability that a person does not pass through the system and is without any security problems, we can use the formula:

Probability = (1 - probability of being a security hazard) * (1 - probability of the security system denying a person without security problems)

Probability = (1 - 0.02) * (1 - 0.015)

Probability = 0.98 * 0.985

Probability = 0.9653

Therefore, the probability that a random person does not pass through the system and is without any security problems is 0.965, or 96.5% (rounded to 3 decimal places).

User Marek Pavelek
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7.7k points
4 votes

Answer:

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

User Avec
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