Question

a certain disease has an incidence rate of 0.1%. if the false negative rate is 7% and the false positive rate is 4%, compute the probability that a person who tests positive actually has the disease. give your answer accurate to at least 3 decimal places question help: video message instructor

Explanation:

Step1: Define the probabilities

Let $P(D)$ be the probability of having the disease. $P(D)=0.001$. The probability of not having the disease is $P(
eg D)=1 - 0.001 = 0.999$. The false - negative rate $P(-|D)=0.07$, so the true - positive rate $P(+|D)=1 - 0.07 = 0.93$. The false - positive rate $P(+|
eg D)=0.04$.

Step2: Use Bayes' theorem

Bayes' theorem states that $P(D|+)=\frac{P(+|D)P(D)}{P(+|D)P(D)+P(+|
eg D)P(
eg D)}$.
Substitute the values:

$$
LATEXBLOCK0
$$

Answer:

$0.023$