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 Virus | Don't Have Virus | Total | | Positive | 9 | 599 | 608 | | Negative | 1 | 5391 | 5392 | | Total | 10 | 5990 | 6000 |
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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