Question

the probability of a false - positive test is 0.10, and the probability of a false - negative (no indication of cancer when cancer is actually present) is 0.20. let (c =) event male patient has prostate cancer (+ =) positive psa test for prostate cancer (- =) negative psa test for prostate cancer a. what is the probability that the male patient has prostate cancer if the psa test comes back positive (to 4 decimals)? b. what is the probability that the male patient has prostate cancer if the psa test comes back negative (to 4 decimals)? c. for older males, the prior probability of having cancer increases. suppose that the prior probability of the male patient is 0.3 rather than 0.02. what is the probability that the male patient has prostate cancer if the psa test comes back positive (to 4 decimals)? what is the probability that the male patient has prostate cancer if the psa test comes back negative (to 4 decimals)? d. what can you infer about the psa test from the results of parts (a), (b), and (c)? the difference between (p(c|+)) and (p(c| -)) in parts (a) and (b) is lower than the difference between (p(c|+)) and (p(c| -)) in part (c).

Explanation:

Step1: Recall Bayes' theorem

$P(C|+)=\frac{P(+|C)P(C)}{P(+|C)P(C)+P(+|\overline{C})P(\overline{C})}$ and $P(C|-)=\frac{P(-|C)P(C)}{P(-|C)P(C)+P(-|\overline{C})P(\overline{C})}$. Assume $P(+|\overline{C}) = 0.10$ (false - positive rate) and $P(-|C)=0.20$ (false - negative rate). Also, $P(\overline{C})=1 - P(C)$.

Step2: Solve part a

Let $P(C) = 0.02$. Then $P(\overline{C})=0.98$. $P(+|C)=1 - P(-|C)=0.80$.
$P(C|+)=\frac{0.80\times0.02}{0.80\times0.02 + 0.10\times0.98}=\frac{0.016}{0.016+0.098}=\frac{0.016}{0.114}\approx0.1404$

Step3: Solve part b

$P(C|-)=\frac{0.20\times0.02}{0.20\times0.02+0.90\times0.98}=\frac{0.004}{0.004 + 0.882}=\frac{0.004}{0.886}\approx0.0045$

Step4: Solve part c

Let $P(C)=0.3$. Then $P(\overline{C}) = 0.7$.
$P(C|+)=\frac{0.80\times0.3}{0.80\times0.3+0.10\times0.7}=\frac{0.24}{0.24 + 0.07}=\frac{0.24}{0.31}\approx0.7742$
$P(C|-)=\frac{0.20\times0.3}{0.20\times0.3+0.90\times0.7}=\frac{0.06}{0.06+0.63}=\frac{0.06}{0.69}\approx0.0870$

Step5: Analyze part d

As the prior probability of having cancer ($P(C)$) increases, the difference between $P(C|+)$ and $P(C|-)$ becomes larger. This indicates that the PSA test is more discriminatory (better at distinguishing between those with and without cancer) when the prior probability of having cancer is higher.

Answer:

a. $0.1404$
b. $0.0045$
c. Positive: $0.7742$, Negative: $0.0870$
d. As the prior probability of having cancer increases, the PSA test is more discriminatory.