Bayes formula help!

Question

Suppose a test for diagnosing a certain disease is successful in detecting the disease in $95$% of all persons infected, but it incorrectly diagnoses $4$% of all healthy people as having the serious disease. Suppose also that it incorrectly diagnoses $12$% of all people having another minor disease as having the serious disease. It is known that $2$% of the population has the serious disease, $90$% of the population is healthy, and $8$% has the minor disease.

Use Bayes formula to find the probability that a person selected at random has the serious disease, given that the test indicates that he or she does Use $H$ to represent healthy, $M$ to represent having the minor disease and D to represent having the serious disease.

This is what I've started with and I'm not sure if I'm on the right track because I don't fully understand this:

$5$% diseased people ($100-95$% successful diagnosis), $96$% of healthy people ($100-4$%incorrectly diagnosed), $88$% minor ($100-12$% of people having minor disease)

$P[M | H] 88 \cdot 8=7.04$%

$P[D | H]\ 5 \cdot 2=0.1$%

$P[H | H]\ 96 \cdot 90=86.4$%

$P[D| H]\ = 0.1+86.4+7.04=93.54$%

I don't understand Bayes Formula at all, please explain... am I on the right track?

Answer 1

I won't work the problem completely, but will try to give you a useful way to start.

You know $P(D) = 0.02, P(T|D) = 0.95,$ and $P(N|D^c) = 0.96,$ where $T$ indicates a positive test (disease detected) and $N = T^c$ indicates a negative test (no detection). You are asked to find $P(D|T).$

It seems to me that the information about the minor disease is a distraction.

If this is correct, then $P(D|T) = P(D\cap T)/P(T),$ and you can find $P(D\cap T) = P(D)P(T|D).$ And then you can find $P(T) = P(D\cap T) + P(D^c\cap T).$

What remains is for you to find $P(D^c\cap T)$ in a manner somewhat similar to the that used to find $P(D\cap T),$ but using complement rules.

Among the "Related" links in the right-hand margin of this page, you should find several completed computations of this type. This section from Wikipedia may also be helpful; let 'User' be $D.$ (This article is not easy to find via Google using sensible search words because commercial junk takes over their first few pages.)

Answer 2

Suppose a test for diagnosing a certain disease is successful in detecting the disease in 95% of all persons infected, but it incorrectly diagnoses 4% of all healthy people as having the serious disease. Suppose also that it incorrectly diagnoses 12% of all people having another minor disease as having the serious disease. It is known that 2% of the population has the serious disease, 90% of the population is healthy, and 8% has the minor disease.

The key here is that you have two random variables.   There is the actual state of the subject, and the result of the test.

Let $S$ be the subject's status, $T$ be the test result; each supporting values $H,M,D$ (healthy, minor disease, or serious disease; as suggested).

List what has been told:

$$\begin{align}\mathsf P(T{=}D \mid S{=}D)&=\phantom{0.95}\\ \mathsf P(T{=}D\mid S{=}H)&=\phantom{0.04}\\ \mathsf P(T{=}D\mid S{=}M)&=\phantom{0.12}\\\mathsf P(S{=}D)&=\phantom{0.02}\\\mathsf P(S{=}H)&=\phantom{0.90}\\\mathsf P(S{=}M)&=\phantom{0.08}\end{align} $$

Use Bayes formula to find the probability that a person selected at random has the serious disease, given that the test indicates that he or she does

Find $\mathsf P(S{=}D\mid T{=}D)$ using Bayes' formula. $$\mathsf P(S{=}D\mid T{=}D)=\dfrac{\mathsf P(T{=}D\mid S{=}D)~\mathsf P(S{=}D)}{\mathsf P(T{=}D)}$$

Also find $\mathsf P(T{=}D)$ using the Law of Total Probability. $$\mathsf P(T{=}D)=\phantom{\ldots}$$