A Canadian male has recently had a Prostate Specific Antigen (PSA) test as to determine if he has prostate cancer. The false-positive rate of a PSA test is $15$%. If he does have prostate cancer, PSA test will be positive $81$% of the time.
Because this male is showing symptoms that are consistent with prostate cancer, it is assumed that the chance he has prostate cancer prior to taking the PSA test is $0.13$.
Part (a) What is the probability that the PSA test will yield a positive result?
Part (b) If the PSA test gives a positive result, what is the probability that he does not have prostate cancer?
Part (c) Suppose the PSA test result is negative, indicating that he does not have prostate cancer and his symptoms are a result of something else. What is the probability that he does have prostate cancer?
I've managed to figure out a), just by adding the probabilities of a positive by drawing a tree. However i'm stuck on the conditional probability part. If I let $A$ be the event that the patient has cancer, and $B$ be the event of a positive test, my attempt for b) was $P(A'|B)$ = $P(A' \cap B) / P(B)$, But I think i'm doing the calculation wrong. Any help would be appreciated
Best Answer
Let $A$ be the event that the patient has cancer, and $B$ be the event of a positive test.
Then $$ P(A'\mid B)=\frac{P(A'\cap B)}{P(B)}=\frac{P(B\mid A')P(A')}{P(B)}=\frac{P(B\mid A')P(A')}{P(B\mid A)P(A)+P(B\mid A')P(A')} $$ by Bayes' rule and the law of total probability. We are given that $$ P(A)=0.13;\quad P(A')=0.87;\quad P(B\mid A')=0.15;\quad P(B\mid A)=0.81 $$ from which you can compute.