Question

Consider the following domain: . a % of men at age fifty who participate in the routine screen have prostate cancer. .b% of men with prostate cancer will get positive prostate-specific antigen test. • % of men without prostate cancer will also get positive prostate-specific antigen test. 1. Please describe the domain knowledge in terms of probabilities in which C is used to denote prostate cancer and P is used to denote positive prostate-specific antigen test. 2. Please follow Bayesian Reasoning (show mathematic steps) to find out if a man in this age group gets a negative prostate-specific antigen test, what is the probability that he doesn't prostate cancer?

Answer 1

1)

a% of men at age fifty have prostate cancer, i.e, probability of having prostate cancer is a/100.

$Pr(C)=a/100$

b% of men with prostate cancer will get positive test, i.e, probability of positive test for men with prostate cancer is b/100.

$Pr(P|C)=b/100$

c% of men without prostate cancer will get positive test, i.e, probability of positive test for men without prostate cancer is c/100.

$Pr(P|\sim C)=c/100$

where ~ denotes not

2)

To find:

probability of men not having prostate cancer given that he tests negative, i.e,

$Pr(\sim C| \sim P)$

i)  

Pr(C) = 1- Pr(C)=1-a/100

ii) $Pr(\sim P| C) = 1 - Pr(P|C) = 1-b/100$

iii) $Pr(\sim P| \sim C) = 1 - Pr(P|\sim C) = 1-c/100$

$Pr(\sim P) = Pr(\sim P, C) + Pr(\sim P, \sim C)$

$= Pr(\sim P|C)*Pr(C)+Pr(\sim P| \sim C)*Pr(\sim C)$

Using 1), ii) and iii),

iv) $Pr(\sim P)= (1-b/100)*a/100+(1-c/100)*(1-a/100)$

$Pr(\sim C| \sim P)=\frac{Pr(\sim C, \sim P)}{Pr(\sim P)}=\frac{Pr(\sim P| \sim C)*Pr(\sim C)}{Pr(\sim P)}$

using i), iii) and iv),

$Pr(\sim C| \sim P)=\frac{(1-c/100)*(1-a/100)}{(1-b/100)*a/100+(1-c/100)*(1-a/100)}$

$Pr(\sim C| \sim P)=\frac{(100-c)*(100-a)}{(100-b)*a+(100-c)*(100-a)}$