Question

According to the CDC and NIH, approximately 1 in 10,000 people have Huntington's disease. Suppose a new test for Huntington's disease is developed. The probability of a false positive is 0.001, and the probability of a false negative is 0.006. If a randomly selected individual is tested using this new test and the result is positive, what is the probability that the individual actually has Huntington's disease? Is this test good enough to consider using?

Answer 1

Ans:

P(disease)=1/10000=0.0001

P(no disease)=1-0.0001=0.9999

P(positive/no disease)=0.001

So,P(negative/no disease)=1-0.001=0.999

P(negative/disease)=0.006

So,P(positive/disease)=1-0.006=0.994

Now,

P(positive)=P(positive/disease)*P(disease)+P(positive/no disease)*P(no disease)

=0.994*0.0001+0.001*0.9999

=0.0011

Using Bayes rule:

P(disease/positive)=P(positive/disease)*P(disease)/P(positive)

=0.994*0.0001/0.0011

=0.0904

No,as probability is very low that the person is actually having disease if found positive.