Re: A probability question

n = normal

s = sick

+ = test positive

- = test negative

Pr(n+) = .1 * .99 = .099

Pr(n-) = .90 * .99 = .891

Pr(s+) = .95 * .01 = .0095

Pr(s-) = .05 * .01 = .0005

Pr(s|+) = .0095/(.099+.0095) = .08756 (=8.756%)

Re: A probability question

Quote:

Originally Posted by

**bryce09** Suppose there is a medical diagnostic test for a disease. The sensitivity of the

test is .95. This means that if a person has the disease, the probability that the

test gives a positive response is .95. The specificity of the test is .90. This

means that if a person does not have the disease, the probability that the test

gives a negative response is .90, or that the false positive rate of the test is .10.

In the population, 1% of the people have the disease.

What is the probability that a person tested has the disease, given the results of the test is positive? Let

D be the event "the person has the disease" and let T be the event "the test

gives a positive result."

I think that I answered this elsewhere. Let $\displaystyle D$ means a person has the disease and $\displaystyle D^c$ means a person does not have the disease. You are given: $\displaystyle P(+|D)=0.95,~P(-|D^c)=0.90,~\&~P(D)=0.01$.

From that we conclude $\displaystyle P(-|D)=0.05,~P(+|D^c)=0.10,~\&~P(D^c)=0.99~.$

Now $\displaystyle P(+)=P(+|D)P(D)+P(+|D^c)P(D^c)$

The question is $\displaystyle P(D|+)=\frac{P(D\cap +)}{P(+)}~.$

Re: A probability question

Thanks for those answers, both were really helpful