False positive and False negative probability using Bayes

Question

Given this problem:

Suppose that a certain disease is present in 10% of the population and that there is a screening test designed to detect this disease if present. The test does not always work perfectly. Sometimes the test is negative when the disease is present, and sometimes it is positive when the disease is absent. The table below shows the proportion of times that the test produces various results:

a. Find the following probabilities from the table: 𝑃(𝐷),𝑃(𝐷𝑐),𝑃(𝑁∣𝐷𝑐),𝑃(𝑁∣𝐷)

b. Use Bayes’ Rule and the results of part a to find P(D|N).

c. Use the definition of conditional probability to find P(D|N). (Your answer should be the same as the answer to part b.)

d. Find the probability of a false positive, that the test is positive, given that the person is disease-free.

e. Find the probability of a false negative, that the test is negative, given that the person has the disease

I believe that there must be something wrong with the exercise because the book says that the answers should be

P(D)= 0.10 and P(𝐷𝑐)= 0.90 which is okay and I do get then P(N|Dc) = 0.94 I do get this one as well

but then all of rest I get wrong

For P(N|D) I am supposed to get 0.2 which I don't and I am doing everything correctly?

Then for b. and c., I get 0.23 which according to the book is correct

Now for d. I am supposed to get 0.056 and I get 0.064 and for e. agin I am supposed to get 0.20 and I get 0.36

My question is is the problem in the exercise or I am not solving it correctly?

Answer 1

The table would make sense if the $0.22$ was in fact $0.02$

Suppose the table said

                     Test positive P    Test negative N  
Disease present D          0.08              0.02       
Disease absent D^c         0.05              0.85       

then you can say

• $\mathbb P(D)=0.08+0.02=0.1$
• $\mathbb P(D^c)=0.05+0.85 =0.9$
• $\mathbb P(P)=0.08+0.05=0.13$
• $\mathbb P(N)=0.02+0.85=0.87$
• $\mathbb P(N \mid D^c) = \frac{0.85}{0.9}\approx 0.9444$
• $\mathbb P(N \mid D) = \frac{0.02}{0.1}=0.2$, the probability of a false negative, that the test is negative given that the person has the disease
• $\mathbb P(D \mid N) = \frac{0.2 \times 0.1}{0.87}=\frac{0.02}{0.87}\approx 0.0230$ using Bayes theorem or conditional probability
• $\mathbb P(P \mid D^c) = \frac{0.05}{0.9}\approx 0.0556$, the probability of a false positive, that the test is positive given that the person is disease-free.