Answer to Question #343864 in Statistics and Probability for chiisti

Question

The prevalence of a certain type of cancer among men aged 55−𝟔𝟎 is 1 in 100. A blood test will be positive 95% of the time if the cancer is present but it is also positive 4% of the time if the cancer is not present. 2.1. In a routine checkup, 56-year-old men receive a positive blood test. What is the probability that he has the type of cancer? 2.2. What is the probability that a randomly selected 56-year-old men tests negative?

Answer 1

Let $A$ denote the event "cancer is pesent", let $B$ denote the event " blood test is positive".

$$P(A)=0.01, P(B|A)=0.95, P(B|A^C)=0.04$$

2.1.

$$P(A|B)=\dfrac{P(A)P(B|A)}{P(A)P(B|A)+P(A^C)P(B|A^C)}$$ $$=\dfrac{0.01(0.95)}{0.01(0.95)+0.99(0.04)}=0.1935$$

2.2

$$P(B^C)=P(A)P(B^C|A)+P(A^C)P(B^C|A^C)$$ $$=0.01(0.05)+0.99(0.96)=0.9509$$