Suppose that one person in 10,000 people has a rare genetic disease.There is an
ID: 3009224 • Letter: S
Question
Suppose that one person in 10,000 people has a rare genetic disease.There is an excellent test for the disease; 99.9% of people with the disease test positive and only0.02% who do not have the disease test positive.
a) What is the probability that someone who tests positive has the genetic disease?
b) What is the probability that someone who tests positive does not have the genetic
disease?c) What is the probability that someone who tests negative has the disease?
d) What is the probability that someone who tests negative does not have the disease?
Explanation / Answer
1 in every 10,000 people has a rare genetic disease
that is the probability of a person to have the disease is P(genetic disease) = P(G.D) = 1/10000
and the probability that the person does not have the disease is P(N.G.D) = 9999/10000
there is a test to tell whether a person has the disease or not that is if he tests positive or negative for the disease
and the probability that the people who have the disease test positive is P(+ G.D) = 99.9% = 99.9/100 = 0.999
=> and the probability that the people who have the disease test negative is P(- G.D) = 1-0.999 = 0.001
the probability that the person who does not have the disease tests positive is P(+ N.G.D) = .02/100 = .0002
=> the probability that the person who does not have the disease tests negative is P(- N.G.D) = 1-.0002 = .9998
we'll be using the Bayes theorem
a> P(+ | G.D) = P( G.D)*P(+ G.D)/[P( G.D)*P(+ G.D) + P(N.G.D)*P(+ N.G.D)]
= (1/10,000)*0.999/((1/10,000)*0.999 +(9999/10,000)*0.0002) =0.3331
b> P(+ | N.G.D) = P( N.G.D)*P(+ N.G.D)/[P(N.G.D)*P(+N. G.D) + P(G.D)*P(+ G.D)]
= (9999/10000)*0.0002/[(9999/10000)*0.0002 +(1/10,000)*0.999]
= 0.6668
c> P(- | G.D) = 1 - P(- | N.G.D) = 1 - .999999 = .0000001
d> P(- | N.G.D) = P( N.G.D)*P(- N.G.D)/[P(N.G.D)*P(-N. G.D) + P(G.D)*P(- G.D)]
= 9999/10000*(.9998)[9999/10000*(.9998)+ 1/10000*(.001)]
= .999999
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