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A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 85% of the time if the person has the virus and 5% of the time if the person does not have the virus, (This 5% result is called a false positive.) Let A be the event "the person is Infected" and B be the event "the person tests positive", a) Find the probability that a person has the virus given that they have tested positive, l.e. find P(AB). Round your answer to the nearest tenth of a percent and do not include a percent sign. P(AIB)= % b) Find the probability that a person does not have the virus given that they test negative, I.e. find P(A'B'). Round your answer to the nearest tenth of a percent and do not include a percent sign. P(A'B') = ​

Respuesta :

This question is solved using the conditional probability concept.

Using this concept, we find that:

  • a) P(AIB)= 5.3%
  • b) P(A'|B') = 99.9%

First, the concept is presented.

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(A|B) = \frac{P(A \cap B)}{P(B)}[/tex]

In which

P(A|B) is the probability of event A happening, given that B happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(B) is the probability of B happening.

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Question a:

For relation with the formula presented above, I will change events A and B.

Event A: Person is infected.

Event B: Positive test.

Probability of a positive test:

  • 85% = 0.85 out of 1/300 (person has the virus).
  • 5% = 0.05 out of 299/300(person does not have the virus)

Thus:

[tex]P(B) = 0.85\frac{1}{300} + 0.05\frac{299}{300} = \frac{0.85\times1 + 0.05\times299}{300} = 0.0527[/tex]

Probability of a positive test and the person is infected.

85% = 0.85 out of 1/300. Thus:

[tex]P(A \cap B) = \frac{0.85}{300} = 0.0028[/tex]

Desired probability:

[tex]P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.0028}{0.0527} = 0.053[/tex]

0.053*100% = 5.3%, thus:

P(AIB)= 5.3%

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Question b:

Event A: Does not have the virus

Event B: Test negative.

Probability of a negative test:

  • 100% - 85% = 15% = 0.15 out of 1/300 (person has the virus).
  • 100% - 5% = 95% = 0.95 out of 299/300(person does not have the virus)

Thus:

[tex]P(B) = 0.15\frac{1}{300} + 0.95\frac{299}{300} = \frac{0.15\times1 + 0.95\times299}{300} = 0.9473[/tex]

Probability of a negative test and the person is not infected.

0.95 out of 299/300

Thus:

[tex]P(A \cap B) = \frac{0.95\times299}{300} = 0.9468[/tex]

Desired probability:

[tex]P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.9468}{0.9473} = 0.999[/tex]

0.999*100% = 99.9%, so:

P(A'|B') = 99.9%

A similar question can be found at https://brainly.com/question/24275491

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