Level of question:Secondary
A rare disease infected 1 in 1000 people in the population. A test for the disease is accurate 99% of the time when given to an infected person and also when given to a heathy person.
- Fill out a two-tier tree diagram and find the probability of the false positive(i.e the conditional probabily of being healty even when tested postive by the test) Comment on the result?
- John after tested once positive was tested again .Fill out completely a three -tier diagram and find the probabilty of false double-positive(ie. the conditional probabily of healthy even john was tested positive twice) what is the probability of a person being healthy and twice-positively tested?
Thankyou for your kindy help!!
I used the letter I to designate infected, H to designate healthy, P to designate a positive test and N to designate a negative test. The two-tier tree diagram is then
The probability of a false positive is the conditional probabily of being healty even when tested postive by the test, that is Pr(H|P). Thus to find the probability of a false positive you need to calculate:
Pr(H|P) = Pr(H and P)/Pr(P)
From the tree diagram
Pr(H and P) = (999/1000) 0.01 = 0.00999.
You can receive a positive test by either being healthy and receiving a positive test or being infected and receiving a positive test. Thus:
Pr(P) = (999/1000) 0.01 + (1/1000) 0.99 = 0.01098
Hence the probability of a false positive is:
Pr(H|P) = Pr(H and P)/Pr(P) = 0.00999/0.01098 = 0.9098
Now put a third tier on the tree diagram to answer part 2.
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