Conditional probabilities can be useful in diagnosing disease. Suppose that thre

Question

Conditional probabilities can be useful in diagnosing disease. Suppose that three different closely related diseases (A1,ApAg) occur in 25%, 15%, and 12% of the population. In addition, suppose that any one of the three mutually exclusive (disjoint) symptom states (B1, B2, B3) may be associated with each of these diseases. Experience shows that the likelihood P(BjlAi) of having a given symptom state when the disease is present is as shown in the following table. Find the probability of disease A2 given symptom B3. Disease State A2 A3 Symptom State B1 B2 A1 08 18 06 68 12 07 64 B3 08 B4 no svmptoms 68

Explanation / Answer

probability of symptom B3 =P(A1)*P(B3|A1)+P(A2)*P(B3|A2)+P(A3)*P(B3|A3)

=0.25*0.06+0.15*0.07+0.12*0.08=0.0351

therefore probability of disease A2 given B3 =P(A2|B3) =P(A2)*P(B3|A2)/P(B3) =0.15*0.07/0.0351=0.299145

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.