Answer to Question #343707 in Statistics and Probability for Takalani

Question #343707

The prevalence of a certain type of cancer among men aged 55βˆ’πŸ”πŸŽ 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?

1
Expert's answer
2022-05-23T16:20:28-0400

Let AA denote the event "cancer is pesent", let BB denote the event " blood test is positive".


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

2.1.


P(A∣B)=P(A)P(B∣A)P(A)P(B∣A)+P(AC)P(B∣AC)P(A|B)=\dfrac{P(A)P(B|A)}{P(A)P(B|A)+P(A^C)P(B|A^C)}

=0.01(0.95)0.01(0.95)+0.99(0.04)=0.1935=\dfrac{0.01(0.95)}{0.01(0.95)+0.99(0.04)}=0.1935

2.2


P(BC)=P(A)P(BC∣A)+P(AC)P(BC∣AC)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=0.01(0.05)+0.99(0.96)=0.9509


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