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Prostate cancer is the most common type of cancer found in males. As an indicator of whether a male has prostate cancer, doctors often perform a test that measures the level of the prostate-specific antigen (PSA) that is produced only by the prostate gland. Although PSA levels are indicative of cancer, the test is notoriously unreliable. Indeed, the probability that a noncancerous man will have an elevated PSA level is approximately .135, increasing to approximately .268 if the man does have cancer. If, on the basis of other factors, a physician is 70 percent certain that a male has prostate cancer, what is the conditional probability that he has the cancer given that

(a) the test indicated an elevated PSA level?

(b) the test did not indicate an elevated PSA level?

Repeat the preceding calculation, this time assuming that the physician initially believes that there is a 30 percent chance that the man has prostate cancer.

Step by step solution

01

Step1: Given Information (part a)

Events:

A - The person has cancer

C - The person has elevated PSA level

02

Step2: Explanation (part a)

Probabilities:

Before the PSA test: $P\left(A\right)=0.7$

Details of the PSA test:

$$\begin{gathered} P\left(\frac{C}{A}\right)=0.268 \\ P\left(\frac{C}{A^C}\right)=0.135 \end{gathered}$$

$A,A^C$are competing hypothesis, by conditioning whether a man has cancer or not.

From Bayers Formula we have:

localid="1646400208652" $$\begin{gathered} P\left(\frac{A}{C}\right)=\frac{P\left({\displaystyle \frac{C}{A}}\right)P\left(A\right)}{P\left({\displaystyle \frac{C}{A}}\right)P\left(A\right)+P\left({\displaystyle \frac{C}{A^C}}\right)P\left(A^C\right)} \\ =\frac{0.268×0.7}{0.268×0.7+0.135×0.3} \\ =\frac{0.1876}{0.2281} \\ =0.8224 \end{gathered}$$

03

Final Result (part a)

0.8224

04

Step4: Given Information (part b)

$P\left(\frac{A}{C^C}\right)$ can be calculated using this equation:

$P\left(\frac{A}{C^C}\right)=\frac{P\left(A{C}^C\right)}{P\left(C^C\right)}$

05

Step5: Explanation (part b)

Numerator :

$$\begin{gathered} P\left(A{C}^C\right)=P\left(\frac{C^C}{A}\right).P\left(A\right) \\ =\left(1-P\left(\frac{C}{A}\right)\right)P\left(A\right) \\ =0.732×0.7 \\ =0.5124 \end{gathered}$$

Denominator:

$$\begin{gathered} P\left(C^C\right)=1-P\left(C\right)=1-O.2281=0.7719 \\ P\left(\frac{A}{C^C}\right)=\frac{0.5124}{0.7719} \\ =0.6638 \end{gathered}$$

06

Step6: Final Result (part b)

0.6638

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