1. **Problem Statement:** We have a test for cancer detection with the following data: - Total adults tested: 1000 - Prevalence of cancer: 1% (so 10 people have cancer) - Test sensitivity (true positive rate): 97% (test indicates cancer in 97% of those who have it) - Test false positive rate: 1% (test indicates cancer in 1% of those who do not have it) We want to find: - $P(\text{Cancer} | \text{Test indicates cancer})$ - $P(\text{Cancer} | \text{Test does not indicate cancer})$ 2. **Formulas and Rules:** We use Bayes' theorem: $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)} $$ where: - $A$ is the event "person has cancer" - $B$ is the event "test indicates cancer" Also, the complement rule: $$ P(\text{not } A) = 1 - P(A) $$ 3. **Calculate probabilities:** - $P(\text{Cancer}) = 0.01$ - $P(\text{No cancer}) = 0.99$ - $P(\text{Test indicates cancer} | \text{Cancer}) = 0.97$ - $P(\text{Test indicates cancer} | \text{No cancer}) = 0.01$ 4. **Calculate $P(\text{Test indicates cancer})$ using total probability:** $$ P(\text{Test indicates cancer}) = P(\text{Test indicates cancer} | \text{Cancer})P(\text{Cancer}) + P(\text{Test indicates cancer} | \text{No cancer})P(\text{No cancer}) $$ $$ = 0.97 \times 0.01 + 0.01 \times 0.99 = 0.0097 + 0.0099 = 0.0196 $$ 5. **Calculate $P(\text{Cancer} | \text{Test indicates cancer})$ using Bayes' theorem:** $$ P(\text{Cancer} | \text{Test indicates cancer}) = \frac{0.97 \times 0.01}{0.0196} = \frac{0.0097}{0.0196} \approx 0.495 $$ 6. **Calculate $P(\text{Test does not indicate cancer})$:** $$ P(\text{Test does not indicate cancer}) = 1 - P(\text{Test indicates cancer}) = 1 - 0.0196 = 0.9804 $$ 7. **Calculate $P(\text{Test does not indicate cancer} | \text{Cancer})$ and $P(\text{Test does not indicate cancer} | \text{No cancer})$:** $$ P(\text{Test does not indicate cancer} | \text{Cancer}) = 1 - 0.97 = 0.03 $$ $$ P(\text{Test does not indicate cancer} | \text{No cancer}) = 1 - 0.01 = 0.99 $$ 8. **Calculate $P(\text{Cancer} | \text{Test does not indicate cancer})$ using Bayes' theorem:** $$ P(\text{Cancer} | \text{Test does not indicate cancer}) = \frac{P(\text{Test does not indicate cancer} | \text{Cancer})P(\text{Cancer})}{P(\text{Test does not indicate cancer})} = \frac{0.03 \times 0.01}{0.9804} = \frac{0.0003}{0.9804} \approx 0.000306 $$ **Final answers:** - Probability of having cancer given test indicates cancer: $\boxed{0.495}$ - Probability of having cancer given test does not indicate cancer: $\boxed{0.0003}$ (rounded to three decimal places)