A population is divided into three (mutually exclusive and exhaustive) classes a

Question

A population is divided into three (mutually exclusive and exhaustive) classes according to the susceptibility to a certain disease: 10% are classified H (high risk), 2054 M (medium risk) and the wart L (low risk). The probability of having the disease for each claw is as follows: 10%, for people in H, 5% for M and 0.5% for the rest. A clinical test for detecting the disease is 99% accurate for people having the disease (so there is a 1% chance that the test is negative for a person having the disease) and 95% accurate for healthy people (so there is a 5% chance that the test is positive for a person that doesn't have the disease). Compute the probabilities for the following events: A person who is tested positive does not have the disease A person who is tested positive is not high risk and does not have the disease.

Explanation / Answer

1) We must determine what fraction test positive.
P(pos.) =P(H)P(pos.|H)+P(M)P(pos.|M)+P(L)(P(pos.|L)
P(L)= 1 - 10%-20%=70%
P(pos.) =.1*.1+.2*.05+.005*.7 = .0235
We are given that the test is 99% accurate (a positive test) for people having the disease .99 * .0235 = 0.023265 P(pos. test and have disease)
The test is 95% accurate for healthy people, which means 5% of healthy people have a positive test.
(1-.0235)*.05 = 0.048825 P(pos. test and no disease)
Thus, P(pos. test) = P(pos. test and have disease)+P(pos. test and no disease)
= 0.023265 + 0.048825 = 0.07209

P(no disease|pos. test) = P(no disease and pos. test)/P(pos. test) =
.048825/.07209 0.6773


2) A person who is tested positive is not high risk and does not have the disease.

P(not high risk and not having the disease) is .2*.95+.7*.995 = .8865

P(not high risk and not having the disease and testing positive) = .8865 * .05 = .044325

Then P(not high risk and not having the disease|test positive) = .044325/.07209 = 0.6149

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