SOLUTION: 28 A certain virus infects one in every 600 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the

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Question 1210431: 28
A certain virus infects one in every 600 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus. Give your answer as a decimal to 4 decimal places.
(a) Find the probability that a person has the virus given that they have tested positive.
(b) Find the probability that a person does not have the virus given that they have tested negative.

Answer by math_tutor2020(3822)   (Show Source): You can put this solution on YOUR website!

Consider a town of 6000 people.
You can make the value larger, but the final answers will be the same regardless of what you pick. I find it's easier to work with multiples of 600.

Since the virus infects one in every 600 people, (1/600)*(6000) = 10 people will be infected with the virus while the other 6000-10 = 5990 people aren't infected.

Of the 10 people infected, the test will correctly say "positive" 90% of the time.
0.90*10 = 9 infected people test positive.
10-9 = 1 infected person gets a false negative.

The test incorrectly reports false positives 10% of the time, so 0.10*5990 = 599 uninfected people get a false positive.
The other 5990 - 599 = 5391 uninfected people get a proper negative result.

Here's a summary written in a table.
Have VirusDon't Have VirusTotal
Positive9599608
Negative153915392
Total1059906000

I recommend using a spreadsheet.


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Part (a)

"Given they have tested positive" means we focus on the "positive" row.
Ignore everything else.
9 people have the virus out of 608 total.

9/608 = 0.0148026315793 approximately
This rounds to 0.0148

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Part (b)

This time we focus on the "negative" row.
There are 5391 people in this row who don't have the virus out of 5392 people total

5391/5392 = 0.999814540059 approximately
This rounds to 0.9998

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