# Bayes' Theorem Probability

I am a bit confused with the following question, I get that P(T|D) = 0.95 and P(D) = 0.0001 however because i'm unable to work out P(T|~D) i'm struggling to apply the theorem, am i missing something? Also i'm unsure about what to do with the information relating to testing negative when you don't have the disease correctly 95% of the time

You take a test T to tell whether you have a disease D. The test comes back positive. You know that test is 95% accurate (the probability of testing positive when you do have the disease is 0.95, and the probability of testing negative when you don’t have the disease is also 0.95). You also know that the disease is rare, only 1 person is 10,000 gets the disease. What is the probability that you have the disease? How would this change if the disease was more common, say affecting 1 person in 100?

You take a test T to tell whether you have a disease D. The test comes back positive. You know that test is 95% accurate (the probability of testing positive when you do have the disease is 0.95, and the probability of testing negative when you don’t have the disease is also 0.95). You also know that the disease is rare, only 1 person is 10,000 gets the disease.

You have $\mathsf P(T\mid D)=0.95, \mathsf P(T^\complement\mid D^\complement)=0.95, \mathsf P(D)=0.000{,}1$ (or $\mathsf P(D)=0.01$ in the second case).

You should know $\mathsf P(D^\complement)=1-\mathsf P(D)$ and $\mathsf P(T\mid D^\complement)=1-\mathsf P(T^\complement\mid D^\complement)$

And of course, Bayes' Rule: $\mathsf P(D\mid T)=\dfrac{\mathsf P(T\mid D)\mathsf P(D)}{\mathsf P(T)}$

Also the Law of Total Probability $\mathsf P(T)~=~\mathsf P(T\mid D)\mathsf P(D)+\mathsf P(T\mid D^\complement)\mathsf P(D^\complement)$

So you have everything you need to find the probability that you have the disease given a positive test result.

Let $D$ be the event that you have the disease and $T$ be the event that the test is positive.

That information tells you that $$P(T|D)=P(\sim T| \sim D)=0.95$$

Notice that $$P(T|\sim D)+P(\sim T|\sim D) = 1$$

Hence you should be able to compute $P(T| \sim D)$.

• Hi, thank you for your response, i'm not good at maths at all (taking comp sci) so would the answer be 0.05? for P(T|~D)? – Kaan Kolcu Oct 7 '17 at 17:45
• yes, that is correct. – Siong Thye Goh Oct 7 '17 at 17:45