Question

a blood test indicates the presence of a particular disease 99% of the time when the disease is actually present. the same test indicates the presence of the disease 0.8% of the time when the disease is not present. four percent of the population actually has the disease. calculate the probability that a person has the disease, given that the test indicates the presence of the disease. choose the correct answer below. a. 0.838 b. 0.98 c. 0.99 d. 0.042

Explanation:

Step1: Define events

Let $D$ be the event that a person has the disease, and $+$ be the event that the test is positive. We are given $P(D)=0.04$, $P(+|D) = 0.99$, and $P(+|
eg D)=0.008$. Also, $P(
eg D)=1 - P(D)=1 - 0.04 = 0.96$.

Step2: Use the law of total - probability to find $P(+)$

By the law of total - probability, $P(+) = P(+|D)P(D)+P(+|
eg D)P(
eg D)$.
Substitute the known values: $P(+)=(0.99\times0.04)+(0.008\times0.96)$.
$P(+) = 0.99\times0.04+0.008\times0.96=0.0396 + 0.00768=0.04728$.

Step3: Use Bayes' theorem to find $P(D|+)$

By Bayes' theorem, $P(D|+)=\frac{P(+|D)P(D)}{P(+)}$.
Substitute the values: $P(D|+)=\frac{0.99\times0.04}{0.04728}=\frac{0.0396}{0.04728}\approx0.838$.

Answer:

A. 0.838