Question

Hello I would like to ask for some help on this two-part practice question that I have, mainly with the second part as I don't really get it.

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 a positive prostate-specific antigen test.
• c% 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 a 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?

So far I have the following but I don't really know how to convert part two as it was one of a few cases  

1. The domain knowledge contains the following 6 possibilities - I'm pretty sure this is right but would like to check it just in case.

1. P (C) = a%
2. P (Bar_C) = 1- a%
3. P (P | C) = b%
4. P ( Bar_P | C) = 1 - b%
5. P (P | Bar_C ) = c%
6. P ( Bar_P| Bar_C ) = 1 - c%
   
   2. P ( | ) = ? = ? = ? This is the bit that I don't know how to convert and need the most help with. I been looking up stuff on it but can't find it for the inverse case anywhere in my textbook or in lectures/online.
   
   
   Extra info: For some extra ref I'm pretty sure right for when it is for the following question:
   
   
   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 does prostate cancer?
   
   Nvm someone else answered similar one that was unanswered bit ago.
   
   
   P (CLP) P(COP) P(P) P(PNC) P(PNC) + P(PNC) P(PC)*P(C) P(P|C)*P(C)+P(P|C)*P(C) (c%)*(1-a%) (b%)*(a%)+(c%)*(1-a%)
   

Answer 1

Solution for given pollenti Given that, " at & non at age -fifty hare Prostate contour, ie, Portality if hring prostate Concert s 9/100 BIC-0/100 by the men with pultorte Couleur will get pouiliw test, le foto dily of positive test for futen nith potnite Coucour blico Ir (ple) a 1100 C'% of run without prostate cancer cuill get positive tent, 159, Probability of Positive test -foy nuen without Postale Carcent is c/100. Pr(Plac) = 6/100 Where a denotes not 2 To find, Bobability or mien not having-Postate cancer given that he tests negative, cils Pr(~C/~P) Pulae):1-Pr(W) = 1-2/100 ,, Pr (wPle) = l-pr (ple)-1-hlivo lii, Pr (~P/NC) = 1-pr(plonic) = 1.6/100 Pr(ap)= Po(~P, ()1 Po(aupa) = Priaple 9 * Pr(c) + P (oplucha polul) using 1, ii, f,ili, i PrP) = (1-bllov) * 9/100-4 (-6/100) * CV- a/nou) Pr Conclup) = Pr(NC, NP) ly Caplace) x Palac) Prlap) Prep)

esing i, iii, fiv (۱- C /1 ) * (1-8/1o) = ( مم) | (oo ) * 6/100 + ( 1 - C / im ) * ( I - 8 // on // ط - 1) . (م-ع) (100 -) * (100 -۵) (- 100) * (- 00)+8* (ط- iro Hence solved, Thank your