Given: Test correct positive $P(+|D)=0.98$ (sensitivity), correct negative $P(-|
eg D)=0.97$(specificity) → false positive$P(+|
eg D)=0.03$, disease prevalence$P(D)=0.008$.

We compute posterior using Bayes: $P(D|+) = \frac{P(+|D)P(D)}{P(+)}$ where $P(+)=P(+|D)P(D)+P(+|
eg D)P(
eg D)$.

• $P(D)=0.008$, $P(
  eg D)=0.992$
• $P(+|D)=0.98$, $P(+|
  eg D)=1-0.97=0.03$

$P(+) =0.98*0.008 +0.03*0.992 =0.00784+0.02976=0.0376$

(i) $P(D|+) =0.00784/0.0376 \approx 0.2085 \approx 20.9\%$

**(ii) $P(
eg D|+) =1 -0.2085=0.7915 \approx 79.2%$** (or$0.02976/0.0376=0.7915$)

(iii) Diagnosis: Despite positive test, patient most probably does NOT have cancer/disease (about 79% chance healthy) because disease is rare (base rate fallacy). Positive predictive value only ~21%. Recommend further testing (more specific test, follow-up) — do not immediately diagnose cancer. If used as MAP classifier, predict $
eg D$ (no cancer).

Note: This classic low-prevalence paradox shows importance of prior. If prevalence higher, posterior rises sharply.