A rare disease affects 1 in 10,000 individuals in the population. A blood test for the
disease is positive for 95% of people with the disease. However, 0.4% of people without
the disease will also test positive.

Question

A rare disease affects 1 in 10,000 individuals in the population. A blood test for the
disease is positive for 95% of people with the disease. However, 0.4% of people without
the disease will also test positive.

anonymous · Answer

(a) Given that a person (selected at random from the general population) tests positive,
what is the probability that she has the disease?

(b) Now assume that the doctor knows that 1 in 100 individuals with symptom G will
have the rare disease, and he only orders the test when individuals have symptom
G. Given that one of the doctor’s patients tests positive, what is the probability that
she has the disease?

anonymous · Answer

Denote $D$ the event that an individual has the disease, and $\overline{D}$ the event that he/she does not.

Also, denote $T$ the event that the test results in a positive, and $\overline{T}$ for a negative.

You're given that the diseases affects 1 in 10,000, so $P(D)=0.0001$, which means $P(\overline{D})=0.9999$. The probabilities of getting positive tests are also given: $P(T|D)=0.95$ and $P(T|\overline{D})=0.004$.

(a) Here you're finding $P(D|T)$. Apply Bayes' formula:
$$P(D|T)=\frac{P(T|D)P(D)}{P(T|D)P(D)+P(T|\overline{D})P(\overline{D})}$$
(b) Now you're given another event; let's call it $G$, the case where an individual exhibits the symptom, and let's call the complement event $\overline{G}$. You have that $P(D|G)=0.01$ and so $P(\overline{D}|G)=0.99$. The tricky part, for me, is what it is you're supposed to find... Sorry I can't help with this one!