Answer to Question #271524 in Statistics and Probability for umut

Question

The probability that a person has a certain disease is 0.05. Medical diagnostic tests are available to determine whether the person actually has the disease. If the disease is actually​ present, the probability that the medical diagnostic test will give a positive result​ (indicating that the disease is​ present) is 0.88. If the disease is not actually​ present, the probability of a positive test result​ (indicating that the disease is​ present) is 0.03.

If the medical diagnostic test has given a positive result​ (indicating that the disease is​ present), what is the probability that the disease is actually​ present?

Answer 1

Let $D$ denote the event "a person has a certain disease", $T$ denote the event "test gives a positive result".

Given $P(D)=0.05,$ $P(T|D)=0.88, P(T|D')=0.03.$

Then $P(D')=1-P(D)=1-0.05=0.95.$

By the Bayes’ Theorem

$$P(D|T)=\dfrac{P(T|D)P(D)}{P(T|D)P(D)+P(T|D')P(D')}$$

$$=\dfrac{0.88(0.05)}{0.88(005)+0.03(0.95)}=\dfrac{88}{145}\approx0.607$$