- #1
ilyas.h
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im sure i followed it correctly but my answer is unusually small...1 in a thousand people have a disease. A company has discovered a new method for testing for the disease.
If a person has the disease, the test will return a +ve result 99% of the time.
If a person doesn't have the disease, the test will return a +ve result 2% of the time.
what is the probability of a person having the disease, given that they have a +ve result?
components:
D = 0.001 (1 in a 1000...)
P(+ve | D) = 0.99
P(+ve | not D) = 0.02
asked to find: P(D | +ve).
Formula (bayes theorem):
P(D | +ve) =
P(D)P(+ve | D)
-----------------------------------------------------
P(D)P(+ve | D) + P(not D)P(+ve | not D)if you plug in the values you get P(D | +ve) = 0.0472
this is clearly too small, if the result was +ve, you'd expect a substantial amount of the +ve cohort to actually have the disease. I've checked the formula 100 times and everything seems correct.
If a person has the disease, the test will return a +ve result 99% of the time.
If a person doesn't have the disease, the test will return a +ve result 2% of the time.
what is the probability of a person having the disease, given that they have a +ve result?
components:
D = 0.001 (1 in a 1000...)
P(+ve | D) = 0.99
P(+ve | not D) = 0.02
asked to find: P(D | +ve).
Formula (bayes theorem):
P(D | +ve) =
P(D)P(+ve | D)
-----------------------------------------------------
P(D)P(+ve | D) + P(not D)P(+ve | not D)if you plug in the values you get P(D | +ve) = 0.0472
this is clearly too small, if the result was +ve, you'd expect a substantial amount of the +ve cohort to actually have the disease. I've checked the formula 100 times and everything seems correct.